The acceleration will be greater for car B.
What is Acceleration ?Acceleration is a physical quantity that describes the rate at which an object's velocity changes over time. It is defined as the change in velocity per unit of time, and it is a vector quantity, meaning it has both magnitude and direction.
When an object experiences acceleration, its velocity changes either in speed, direction, or both. If the acceleration is in the same direction as the velocity, the object's speed increases. If the acceleration is in the opposite direction as the velocity, the object's speed decreases. If the acceleration is perpendicular to the velocity, the object's direction changes.
The standard unit of acceleration is meters per second squared (m/s^2). A positive acceleration indicates that an object is speeding up, while a negative acceleration indicates that an object is slowing down. If the acceleration is zero, then the object's velocity is constant.
According to the given information:To calculate the acceleration of each car around its respective curve, we can use the following formula:
a = v^2 / r
where a is the centripetal acceleration of the car, v is its speed, and r is the radius of the curve.
For car A, traveling at speed v around curve number one with radius r, the acceleration is:
a_A = v^2 / r
For car B, traveling at speed 2v around curve number two with radius 2r, the acceleration is:
a_B = (2v)^2 / (2r) = 4v^2 / (2r) = 2v^2 / r
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the three forces are applied to the bracket. determine the range of values for the magnitude of force p so that the resultant of the three forces does not exceed 2400 n. Force P is always directed to the right.
Range of values for given condition will depend on magnitude and directions of the three given forces. [tex]|R| < = 2400 N[/tex]
To find the range of values for the magnitude of force P so that the resultant of the three forces does not exceed 2400 N, find magnitude and direction.
We can do this by using vector addition. Adding the three forces together, we get:
R = F1 + F2 + F3
where F1, F2, and F3 are the magnitudes and directions of the three given forces, and R: magnitude with resultant force direction.
To ensure that the resultant force does not exceed 2400 N, we must have:
|R| <= 2400 N
Therefore, we need to find the range of values for the magnitude of force P that satisfy this inequality.
The magnitudes and directions of the three given forces are not specified in the problem, so we cannot provide a specific numerical answer. However, we can provide a general method for solving the problem.
To find the range of values for the magnitude of force P, we can first find the maximum and minimum values of the magnitude of the resultant force for different values of P. We can then find the range of values of P that satisfy the inequality above.
This can be done numerically by using vector addition and trigonometry to find the magnitude and direction of the resultant force for different values of P. Alternatively, we can use graphical methods such as force polygons or vector diagrams to visualize the resultant force and find the range of values of P that satisfy the inequality.
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What is the Reynolds’ number if the average flow speed of blood through the coronary artery is 15 mL/s, the density of the blood is 50 kg/m3 and the vessel has a diameter of 0.2 m?
five coulombs (5 c) of charge pass through the element from point a to point b. if the energy absorbed by the element is 120 j, determine the voltage across the element.
The voltage across the element is 24 V where if the energy absorbed by the element is 120 j.
Given data as per the question:
Charges = 5 coulomb
Energy absorbed by the element = 120 J
As per the formula we have,
Energy = Voltage X Charge
120= Voltage X 5
Voltage = 120/5 = 24 V
The voltage in the whole process will be negative because the energy is absorbed.
In a series circuit, the current is the same for all the components. While the circuit reaches its steady state, the capacitor charges and the voltage across its plates increases until it reaches the one on the terminals, and at that point it is in the steady state.
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If the energy absorbed by the element is 120 j, So 5*V1 =120 , V1 =24 Volts.
The voltage distinction be V1 Volts.
When 5C of charge moves from A to B, its energy increments by 120J.
So 5*V1 =120
V1 =24 Volts.
The voltage distinction is hence 24 Volts.
At the point when the charge moves from higher potential to lower, it loses energy and when it moves from lower potential to higher, it retains energy. The energy ingested (or lost) is relative to the potential (voltage) contrast between the two focuses.
The voltage in the entire cycle will be negative in light of the fact that the energy is retained.
In a series circuit, the current is no different for every one of the parts. While the circuit arrives at its consistent express, the capacitor charges and the voltage across its plates increments until it arrives at the one on the terminals, and by then it is in the consistent state.
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(Note that the strain in this case is uniform.) Also calculate change in distance using geometrically non-linear strain, and compare. [For those with the 1st edition of the text, the last sentence of Prob. 2.3 should be replaced by the following; " Calculate the strain components corresponding to the given displacement field. Use the definition of εxx to estimate the change of distance between the two points. Compare the two results.]
We can estimate the change in distance between the two points: ΔL = εxxL0 = 0.002500mm = 1mm for given displacement field.
In a geometrically linear analysis, the strain is assumed to be proportional to the displacement, which means that the strain is uniform throughout the material. However, in reality, the strain is not always linearly related to the displacement, and a more accurate analysis would take into account the non-linear strain behavior.
To calculate the change in distance using geometrically linear strain, we can use the definition of strain: ε = ΔL/L0, where ΔL is the change in length and L0 is the original length. In this case, we are given that εxx = 0.002, which means that the strain in the x-direction is 0.2%. Using this equation, we can estimate the change in distance between the two points: ΔL = εxxL0 = 0.002500mm = 1mm.
To calculate the change in distance using geometrically non-linear strain, we would need to use a more complex strain-displacement relationship. However, we can expect that the non-linear strain analysis would predict a slightly different change in distance compared to the linear analysis, since the strain would not be assumed to be uniform throughout the material.
Comparing the results from the two analyses, we see that the change in distance predicted by the geometrically linear strain analysis is 1mm. While we cannot accurately predict the change in distance using the non-linear strain analysis without further information, we can expect the difference between the two results to be small. Nonetheless, in situations where more accurate predictions are necessary, a geometrically non-linear analysis may be required.
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Finding an unknown weight
As shown, a mass is hung from the pulley. This mass causes a tensile force of 16.0 N in the cable and the pulley to hang h=130.0 cm from the ceiling. Assume that the pulley has no mass. What is the weight of the mass?
Found W to be 27.7 N
b) Finding the mass of the pulley
As shown, an object with mass m=5.8 kg is hung from a pulley and spring system. When the object is hung, the tension in the cable is 41.8 N and the pulley is h=147.2 cm below the ceiling.
Because the tensile force is greater than the object's weight, the pulley cannot be massless as assumed. Find the mass of the pulley. For this problem, use g
=
9.81
m
/
s
2
. Not 1.95 kg
The unknown weight is 16 N and the mass of the pulley is 4.26 kg.
What is the unknown weight?
The weight of the mass can be found using the formula;
W = T
where;
T is the tension in the cable,so W = 16.0 N.
To find the mass of the pulley, we need to take into account its weight and the tension in the cable. Since the tension is greater than the weight of the object, we know that the tension is also supporting the weight of the pulley.
T - m_pulley x g = 0
where;
m_pulley is the mass of the pulley and g is the acceleration due to gravity.m_pulley = T/g
= 41.8 N/9.81 m/s^2 = 4.26 kg
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A fishing boat in the ocean is moving at a speed of 20.0 km/h and heading in a direction of 40.0° east of north. A lighthouse spots the fishing boat at a distance of 24.0 km from the lighthouse and in a direction of 15.0° east of north. At the moment the fishing boat is spotted, a speedboat launches from a dock adjacent to the lighthouse. The speedboat travels at a speed of 44.0 km/h and heads in a straight line such that it will intercept the fishing boat.
(a)How much time, in minutes, does the speedboat take to travel from the dock to the point where it intercepts the fishing boat?
(b)In what direction does the speedboat travel? Express the direction as a compass bearing with respect to due north.
In order to reach the fishing boat in the smallest amount of time and distance, the speedboat's pilot should point it 15.0 + 13.7 = 28.7° east of true north from the lighthouse.
What is the fishing boat?Typically speaking, if I see the initial line of sight from the lighthouse as being straight north, this is the simplest for me to solve.
The fishing vessel is thus traveling 40.0 – 15.0 = 25.0° east of north.
The fishing boat's eastward speed is 29.0sin25.0, or 12.3 kilometers per hour.
The fishing boat's northward speed is 29.0cos25.0, or 26.3 kilometers per hour.
If the speedboat matches the fishing boat's eastward speed such that the two boats stay on a line that is exactly north/south of one another, the speedboat will go the smallest distance in the quickest time.
The speedboat makes northward progress at
√(52.0² - 12.3²) = 53.4 km/hr
The net northward speed difference is
53.4 – 26.3 = 27.1 km/hr
so the gap between them closes in
16 km/27.1km/hr = 0.6 hr (or 36 min)
The speed boat will be traveling in the direction θ to the right of our artificial north
sinθ = 12.3/52.0
θ = 13.7°
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Two wooden members of uniform rectangular cross section are joined by the simple glued scarf splice shown. Knowing that P= 18 kN, determine the normal and shearing stresses in the glued splice. 150 mm P 75 mm kPa. The normal stress is ... kPaThe shearing stress is ... kPa.
1.12 MPa is the shear stress in the glued joint.
What is class 10 cross section?If you think of an object as a 2D object, its cross section area is one of its areas. Consider a perfectly spherical ball as an illustration. A circle with a radius equal to the ball can be seen if you view the ball as a 2D object. Consider a cone for another illustration.
The axial or normal stress, σ, in the glued joint can be calculated as:
σ = P/A
A = 2 × b × h
A = 2 × 75 mm × 150 mm = 22,500 mm²
Substituting the values of P and A, we get:
σ = 18 kN / 22,500 mm² = 0.8 MPa
So the normal stress in the glued joint is 0.8 MPa.
The shear stress, τ, in the glued joint can be calculated as:
τ = VQ/It
V = P/2 = 9 kN
The first moment of area, Q, can be calculated as:
Q = b×h2/4
When we change the values of b and h, we obtain:
Q = 75 mm × (150 mm)2 / 4 = 1,406,250 mm³
Calculating the second instant of area, I, is as follows:
I = b×h3/12
The result of substituting the values of b and h is:
I = 75 mm × (150 mm)3 / 12 = 1,054,687.5 mm⁴
Substituting the values of V, Q, I, and t, we get:
τ = 9 kN × 1,406,250 mm³/ (1,054,687.5 mm⁴ × 75 mm) = 1.12 MPa
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Calculate the velocity a spherical rain drop would achieve falling (taking downward as positive) from 4.4km in the following situations. (h=4.4km; l=3.8mm; d=1.16kg/m3
a. Calculate the velocity in the absence of air drag in m/s. ____
b. Calculate the velocity wiith air drag in m/s Take the size across of the frop to be 3.8mm, the dnesity of air to be 1.16kg/m3 , he density of water to be 1000kg/m3 , the surface area to be\pir2, and the drage coefficient to be 1.0. _____
To calculate the velocity of a spherical raindrop falling from 4.4 km without air drag, we can use the equations of motion:
v² = u² + 2as
How to calculate velocity?where v is the final velocity, u is the initial velocity (which is 0 in this case), a is the acceleration due to gravity, s is the displacement (which is equal to the height of the fall), and we assume that downward is positive. We can use the acceleration due to gravity as -9.81 m/s². Therefore, we get:v² = 0 + 2(-9.81 m/s²)(4.4 km) = -2(9.81 m/s²)(4400 m) = -86,140 m²/s²
Since the velocity cannot be negative, we take the square root of the magnitude to get the final velocity:v = sqrt(86,140 m²/s²) = 293.9 m/s
Therefore, the velocity of the raindrop falling from 4.4 km without air drag is approximately 293.9 m/s.
b. To calculate the velocity of the raindrop falling with air drag, we can use the following equation:
F_drag = 1/2 * rho * v^2 * A * C_d
where F_drag is the drag force, rho is the density of air, v is the velocity of the raindrop, A is the cross-sectional area of the raindrop (which is pi*(d/2)^2), and C_d is the drag coefficient. We can assume that the weight of the raindrop is balanced by the upward force due to air resistance, so we can write:F_drag = m * g
where m is the mass of the raindrop (which is (4/3)pi(l/2)^3*1000 kg/m³), and g is the acceleration due to gravity. We can rearrange the two equations to get:m * g = 1/2 * rho * v^2 * A * C_d
Solving for v, we get:
v = sqrt((2 * m * g) / (rho * A * C_d))
Substituting the values, we get:
v = sqrt((2 * (4/3) * pi * (3.8/2)^3 * 1000 kg/m³ * 9.81 m/s²) / (1.16 kg/m³ * pi * (3.8/2)^2 * 1.0))
Simplifying, we get:
v = 9.7 m/s
Therefore, the velocity of the raindrop falling from 4.4 km with air drag is approximately 9.7 m/s.
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A jet of water emerging from a hole in the side of a tanke of water covers horizontal distance R before Striking the ground. If the depth of water in the tank is h and the height of the hole from bottom of the tank is yo formula for R for an Identical the derive a fank on the moon where Jm = ¼ де R? Show that the maximum range of jet of water is Rmax = hand is achieved when =h/2 y what is the
The maximum range of the water jet is[tex]Rmax = h(sqrt(2)),[/tex] and it is achieved when the height of the hole is [tex]yo = h/2.[/tex]
What is range?We can use the equations of motion for an object under constant acceleration to derive the formula for the horizontal distance R that a jet of water will travel before striking the ground. The acceleration of the water jet is due to gravity, and it is constant and equal to the acceleration due to gravity, g.
Let t be the time it takes for the water jet to hit the ground after leaving the hole. We can use the equation of motion for the vertical direction:
[tex]y = yo + voy t - (1/2)gt^2[/tex]
where y is the vertical displacement of the water jet, yo is the initial vertical displacement (the height of the hole from the bottom of the tank), voy is the initial vertical velocity (which is zero), and g is the acceleration due to gravity.
Solving for t, we get:
[tex]t = sqrt((2(yo - y))/g)[/tex]
Now we can use the equation of motion for the horizontal direction:
[tex]x = v_{0} x t[/tex]
where x is the horizontal displacement (which is R), and vox is the initial horizontal velocity (which is constant and equal to the velocity of the water jet as it emerges from the hole).
We can express [tex]v_{0} x[/tex] in terms of the vertical displacement y and the time t:
[tex]v_{0} x = R/t = R(sqrt(g/(2(yo - y))))[/tex]
Substituting for t and simplifying, we get:
[tex]v_{0} x = R(sqrt(2g/h))[/tex]
Now we can express the range R in terms of the tank height h and the height of the hole yo:
[tex]R = (v_{0} x^2/h) = 2h(sqrt(yo/h))[/tex]
To derive the formula for an identical tank on the moon where the acceleration due to gravity is Jm = 1/4 of the acceleration due to gravity on Earth (g), we can substitute g/4 for g in the equation for the horizontal velocity vox. This gives:
[tex]vox = R(sqrt(g/(8h)))[/tex]
Substituting into the equation for R, we get:
[tex]R = (vox^2/h) = 8h(sqrt(yo/h))[/tex]
To show that the maximum range of the water jet is achieved when yo = h/2, we can differentiate R with respect to yo and set the result equal to zero:
[tex]dR/dyo = (4/h)(sqrt(yo/h))(h/2 - yo)[/tex]
Setting this equal to zero and solving for yo, we get:
[tex]yo = h/2[/tex]
To find the maximum range Rmax, we substitute yo = h/2 into the equation for R:
[tex]Rmax = 2h(sqrt(h/2h)) = h(sqrt(2))[/tex]
Therefore, the maximum range of the water jet is [tex]Rmax = h(sqrt(2))[/tex], and it is achieved when the height of the hole is yo = h/2.
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List the homologous series
The organic compounds in the homologous series have similar chemical properties. The simplest example of homologous series is the first four hydrocarbons; methane, ethane, propane and butane.
What is homologous series?The homologous series is known as the group of organic compounds that differ from each other by a methylene group. They are series of compounds with the same functional group and similar chemical properties.
The compounds of carbon in homologous series have different number of carbon atoms. But they contain the same functional group. Alkanes, alkenes and alkynes form the homologous series.
Thus all the alkanes, alkenes and alkynes form homologous series.
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which of the following would be needed to calculate the rate in units of concentration per time? which of the following would be needed to calculate the rate in units of concentration per time? the pressure of the gas at each time the molecular weight of a the temperature the volume of the reaction flask
The volume of the reaction flask would be needed to calculate the rate in units of concentration per time. Option d is correct.
Reaction rates are usually expressed as the concentration of reactant consumed or the concentration of product formed per unit time. The units are thus moles per liter per unit time, written as M/s, M/min, or M/h.
Chemists start a reaction, measure the reactant or product concentration at various points as the reaction advances, and maybe display the concentration as a function of time on a graph. Then they compute the change in concentration per unit time.
The volume of the reaction flask is required to know the concentration of the solution. Thus the rate.
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--The complete question is, Which of the following would be needed to calculate the rate in units of concentration per time?
a. the pressure of the gas at each time
b. the molecular weight of a
c. the temperature
d. the volume of the reaction flask--
Two identical conducting small spheres are placed with their centers 0.300m apart. One is given a charge of 12.0nC and the other a charge of −18.0nC.
(a) Find the electric force exerted by one sphere on the other.
(b) What If? The spheres are connected by a conducting wire. Find the electric force each exerts on the other after they have come to equilibrium.
Electric force by one sphere on other is [tex]-4.80 * 10^(-3) N[/tex] and after they have come to equilibrium is [tex]1.08 * (10^-3) N[/tex]
(a) The electric force exerted by one sphere on the other can be calculated using Coulomb's law:
F = k * (q1 * q2) / r^2
where F is the force, k is Coulomb's constant (9.0 x 10^9 Nm^2/C^2), q1 and q2 are the charges on the two spheres (12.0nC and -18.0nC), and r is the distance between their centers (0.300m).
Plugging in the values, we get:
F = 9.0 x 10^9 * (12.0 x 10^-9) * (-18.0 x 10^-9) / (0.300)^2 = -4.80 x 10^-3 N
The negative sign indicates that the force is attractive.
(b) When the spheres are connected by a conducting wire, they will share charge until they reach equilibrium. At equilibrium, the net force on each sphere will be zero. The electric force each sphere exerts on the other will be the same in magnitude, but opposite in direction.
To find the magnitude of the force, we can use the fact that the potential of each sphere will be the same at equilibrium. The potential can be calculated using:
V = k * q / r
where V is the potential, k is Coulomb's constant, q is the charge on the sphere, and r is the distance from the sphere's center.
At equilibrium, the potential of each sphere will be the same, so:
k * q1 / r1 = k * q2 / r2
Solving for q1 and q2, we get:
q1 = -q2 * r1 / r2
Plugging in the values, we get:
q1 =[tex]-(-18.0 * 10^(-9)) * 0.150m / 0.150m = 18.0 * 10^(-9) C[/tex]
q2 = -q1 = [tex]-18.0 * 10^-(9) C[/tex]
The electric force each sphere exerts on the other can be calculated using Coulomb's law with the new charges:
F = k * (q1 * q2) / r^2
Plugging in the values, we get:
F =[tex]9.0 * 10^(-9) * (18.0 * 10^(-9)^2 / (0.300)^2 = 1.08 * 10^(-3) N[/tex]
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a crate of mass 190 kg is being pulled along a horizontal, flat surface by a massless rope. the surface has known coefficients of friction, where the coefficient of kinetic friction is y?
If you keep pulling with the same tension, what is the acceleration of the crate?
The rope's tension, the degree of friction factor, and also the mass of crate all affect the crate's acceleration.
How does friction work in simple physics?Between contacting materials that are sliding or attempting to slide over one another, there's an external force called friction. For illustration, friction makes it challenging to push a book down the floor.
Net force is equal to the sum of the resistive and rope forces.
The applied load is the same as the weight of crate, or mg, with g is the speed from gravity when m is the container's mass because the floor is straight and the object is not moving upwards.
In this case, the amount of contact is given by:
frictional force = y * mg
where y represents the kinetic friction coefficient.
When everything is added, we have:
net force is equal to T-y*mg.
Using Newton ’s second as well:
T-y*mg = ma
where a represents the crate's acceleration.
Solving for a, we get:
a = (T - y * mg) / m
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**if ur really good at this stuff, lmk in the comments and ill be willing to pay u to tutor me in this stuff
1- A crate with a mass of 42.5 kg is pushed with a horizontal force of 150 N. The crate moves at a constant speed across a level, rough floor a distance of 4.80 m.
(a) What is the work done (in J) by the 150 N force? (J)
(b) What is the coefficient of kinetic friction between the crate and the floor?
2- Starting from rest, a 4.60-kg block slides 2.40 m down a rough 30.0° incline. The coefficient of kinetic friction between the block and the incline is k = 0.436.
(a) Determine the work done by the force of gravity. (J)
(b) Determine the work done by the friction force between block and incline. (J)
(c) Determine the work done by the normal force. (J)
(d) Qualitatively, how would the answers change if a shorter ramp at a steeper angle were used to span the same vertical height?
4- A 0.46-kg particle has a speed of 5.0 m/s at point A and kinetic energy of 8.5 J at point B.
(a) What is its kinetic energy at A? (J)
(b) What is its speed at point B? (m/s)
(c) What is the total work done on the particle as it moves from A to B?
5- A man pushing a crate of mass m = 92.0 kg at a speed of v = 0.870 m/s encounters a rough horizontal surface of length ℓ = 0.65 m as in the figure below. If the coefficient of kinetic friction between the crate and rough surface is 0.350 and he exerts a constant horizontal force of 289 N on the crate. A man pushes a crate labeled m, which moves with a velocity vector v to the right, on a horizontal surface. The horizontal surface is textured from the right edge of the crate to a horizontal distance ℓ from the right edge of the crate.
(a) Find the magnitude and direction of the net force on the crate while it is on the rough surface.
(b) Find the net work done on the crate while it is on the rough surface. (J)
(c) Find the speed of the crate when it reaches the end of the rough surface. (m/s)
6- A block of mass 3.80 kg is placed against a horizontal spring of constant k = 895 N/m and pushed so the spring compresses by 0.0650 m.
HINT
(a) What is the elastic potential energy of the block-spring system (in J)? (J)
(b) If the block is now released and the surface is frictionless, calculate the block's speed (in m/s) after leaving the spring. (m/s)
7- A child on a sled with a total mass of 46.0 kg slides down an icy hillside with negligible friction. The sled starts from rest and has a speed of 3.30 m/s at the bottom. What is the height of the hill (in m)?
8- The figure below shows a box with a mass of m = 7.20 kg that starts from rest at point A and slides on a track with negligible friction. Point A is at a height of ha = 6.90 m.
An illustration shows a wavy track, starting from a crest, moving to a trough, then again to a crest and trough, and finally to a crest that then moves downward. Three points in the track are highlighted, A, B, and C. Point A is at the top of the track where a box of mass m is placed ready to get released. It is at the height labeled ha from the ground. Point B is shown at the next crest and is at a height of 3.20 meters from the ground. Point C is shown at the following trough and is at a height of 2.00 meters from the ground.
(a) What is the box's speed at point B (in m/s)? (m/s) What is the box's speed at point C (in m/s)?
(b) What is the net work (in J) done by the gravitational force on the box as it moves from point A to point C? (J)
9- A superhero swings on a 35.0 m long cable initially inclined at an angle of 35.0° with the vertical. (Assume the cable has negligible mass.)
(a) What is the superhero's speed (in m/s) at the bottom of the swing if he starts from rest? m/s
(b) What is the superhero's speed (in m/s) at the bottom of the swing if instead he pushes off with a speed of 5.00 m/s?
(a) The work done by the 150 N force is given by W = F * d * cos(Θ), where Θ is the angle between the force and displacement vectors and d is the distance over which the force is applied. In this case, the angle is 0° (the force is applied in the same direction as the displacement), so the work done is simply W = F * d = 150 N * 4.80 m = 720 J.
What are the responses to other questions?(b) The coefficient of kinetic friction can be calculated using the equation F_friction = μ_k * F_norm, where F_friction is the force of friction, μ_k is the coefficient of kinetic friction, and F_norm is the normal force. The normal force is equal to the weight of the crate, which is given by F_norm = m * g, where m is the mass of the crate and g is the acceleration due to gravity.
So, we have:
F_friction = μ_k * m * g
150 N = μ_k * 42.5 kg * 9.8 m/s^2
Solving for μ_k, we get μ_k = 150 N / (42.5 kg * 9.8 m/s^2) = 0.0313.
2. a) The work done by the force of gravity can be calculated using the formula:
work_gravity = m * g * cos(θ) * d
where m is the mass of the block (4.60 kg), g is the acceleration due to gravity (9.8 m/s^2), θ is the angle of incline (30.0°), and d is the distance traveled (2.40 m).
cos(30°) = 0.866025
work_gravity = 4.60 kg * 9.8 m/s^2 * 0.866025 * 2.40 m = 68.44 J
b) The work done by the friction force can be calculated using the formula:
work_friction = -friction_force * d
where friction_force is the force of friction between the block and incline, which can be calculated as:
friction_force = k * normal_force
where k is the coefficient of kinetic friction (0.436) and normal_force is the normal force exerted on the block, which can be calculated as:
normal_force = m * g * sin(θ)
sin(30°) = 0.5
normal_force = 4.60 kg * 9.8 m/s^2 * 0.5 = 22.47 N
friction_force = 0.436 * 22.47 N = 9.82 N
work_friction = -9.82 N * 2.40 m = -23.58 J
c) The work done by the normal force is zero since the normal force is perpendicular to the direction of motion and does not perform any work.
4. a) The kinetic energy of a particle can be calculated using the formula:
KE = 0.5 * m * v^2
where m is the mass of the particle (0.46 kg) and v is its velocity (5.0 m/s).
KE_A = 0.5 * 0.46 kg * (5.0 m/s)^2 = 12.75 J
b) The kinetic energy of a particle can be used to find its velocity using the formula:
KE = 0.5 * m * v^2
Rearranging this equation to solve for velocity:
v = sqrt(2 * KE / m)
v_B = sqrt(2 * 8.5 J / 0.46 kg) = 4.76 m/s
c) The work done on the particle as it moves from A to B can be calculated as the change in its kinetic energy:
work = ΔKE = KE_B - KE_A = 8.5 J - 12.75 J = -4.25 J
4. a) The kinetic energy of a particle can be calculated using the formula:
KE = 0.5 * m * v^2
where m is the mass of the particle (0.46 kg) and v is its velocity (5.0 m/s).
KE_A = 0.5 * 0.46 kg * (5.0 m/s)^2 = 12.75 J
b) The kinetic energy of a particle can be used to find its velocity using the formula:
KE = 0.5 * m * v^2
Rearranging this equation to solve for velocity:
v = sqrt(2 * KE / m)
v_B = sqrt(2 * 8.5 J / 0.46 kg) = 4.76 m/s
c) The work done on the particle as it moves from A to B can be calculated as the change in its kinetic energy:
work = ΔKE = KE_B - KE_A = 8.5 J - 12.75 J = -4.25 J
The negative sign indicates that the work done on the particle was done against its motion, decreasing its kinetic energy.
5. a) The net force on the crate can be found by considering the horizontal forces acting on it. These forces include the applied force F_applied and the frictional force F_friction:
F_friction = friction_coefficient * normal_force
where friction_coefficient is the coefficient of kinetic friction (0.350) and normal_force is the normal force acting on the crate, which can be calculated as:
normal_force = m * g
where m is the mass of the crate (92.0 kg) and g is the acceleration due to gravity (9.8 m/s^2).
normal_force = 92.0 kg * 9.8 m/s^2 = 903.6 N
F_friction = 0.350 * 903.6 N = 317.3 N
The net force is given by:
F_net = F_applied - F_friction = 289 N - 317.3 N = -28.3 N
The negative sign indicates that the net force is in the direction opposite to the direction of the applied force, i.e., to the left.
b) The net work done on the crate can be calculated using the formula:
work = force * distance * cos(θ)
where force is the net force on the crate (-28.3 N), distance is the distance traveled along the rough surface (0.65 m), and θ is the angle between the net force and the direction of motion, which is 0° since the net force and the displacement are in opposite directions.
work = -28.3 N * 0.65 m * cos(0°) = -18.4 J
c) The final kinetic energy of the crate can be calculated using the formula:
KE_final = KE_initial + work
where KE_initial is the initial kinetic energy of the crate, which can be calculated as:
KE_initial = 0.5 * m * v^2
KE_initial = 0.5 * 92.0 kg * (0.870 m/s)^2 = 6.66 J
KE_final = 6.66 J - 18.4 J = -11.7 J
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The mass is 18 kg. The velocity is 4.7 m/s. What is the kinetic energy?
The kinetic energy is 4400 joules. The mass is 29 kg. What is the speed?
Answer:
Explanation:
Kinetic energy = 1/2*m*v^2
The kinetic energy is 4400 joules. The mass is 29 kg. What is the speed
4400 = 1/2*29*v^2
v^2 = 303.44
v=17.42m/s^2
The mass is 18 kg. The velocity is 4.7 m/s. What is the kinetic energy?
KE= 1/2*18*4.7*4.7=198.81J
find the work done by moving an object 3 feet from (0,0) to (3,0) by a force of 15lbs in the direction of (4, 1)
The force of 15 lbs in the direction of (4,1) is: 41.3 ft-lb.
The work done by a force of 15lbs moving an object from (0,0) to (3,0) in the direction of (4,1) is given by the formula:
W = Fdcos(θ), where F is the force, d is the distance, and θ is the angle between the force and the direction of motion.
In this case, θ = arctan(1/4) = 14.036 degrees, so the work done is W = 153cos(14.036) = 41.3 ft-lb.
Force is a physical quantity that is defined as the rate of change of momentum. It is a vector, meaning that it has both magnitude and direction. Force is usually represented by the symbol F and is measured in newtons (N).
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The troubleshooting of a parallel circuit that contains three dimly lit bulbs is being discussed. A voltmeter that is placed across each of the bulbs indicates 7 2 volts.
Technician A says that the power supply that is common to all three bulbs may be faulty.
Technician B says that the ground terminal that is common to all three bulbs may have excessive resistance.
Who is correct?
A. A only
B. B only
C. Both A and B
D. Neither A nor B
The correct answer is C. Both A and B are correct, as the dimly lit bulbs could be due to either a faulty power supply or excessive resistance in the ground terminal.
Both technician A and technician B are correct when a voltmeter is placed across each of the bulbs indicates 7 2 volts during the troubleshooting of a parallel circuit that contains three dimly lit bulbs. Technician A says that the power supply that is common to all three bulbs may be faulty. Technician B says that the ground terminal that is common to all three bulbs may have excessive resistance.
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The decay constant for sodium-24, a radioisotope used medically in blood studies, is 4.63x10-2 h-1. What is the t1/2 of 24Na?
The relationship between the decay constant ahd half life of the radioactive isotope is given as :
So putting all the values , we get :
4.63 x 10-2 = 0.693 / (t1/2)
t1/2 = 14.97 hr
So the half life of sodium - 24 is 14.97 hours.
What is radioisotopes ?
a chemical element in an unstable state that emits radiation as it decomposes and becomes more stable. Radioisotopes can be created in a lab or in the natural world. They are utilised in imaging studies and therapy in medicine. likewise known as radioisotopes.
Hence 14.97 hours is a correct answer.
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A piece of wire is cut into two pieces, A and B, which are then tightly stretched and moun rigid walls. A and B have the same stretched lengths, but A is stretched more tightly following quantities will always be larger for waves on A than for waves on B? a) amplitude of the wave b) frequency of the first harmonic c) wave velocity d) wavelength of the first harmonic e) both b and c
The correct answer is e) both b and c.
The quantities that will always be larger for waves on A than for waves on B are the frequency of the first harmonic and the wave velocity.
The frequency of the first harmonic is determined by the tension in the wire. Since A is stretched more tightly than B, the frequency of the first harmonic will be larger for waves on A than for waves on B.
The wave velocity is also determined by the tension in the wire. A higher tension results in a higher wave velocity. Therefore, the wave velocity will also be larger for waves on A than for waves on B.
The amplitude and wavelength of the first harmonic are not affected by the tension in the wire, so they will not be larger for waves on A than for waves on B.
In conclusion, the frequency of the first harmonic and the wave velocity will always be larger for waves on A than for waves on B.
Therefore, the correct answer for quantities that will be larger for waves on A than for waves on B is option e) both b and c.
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A gas is enclosed within a chamber that is fitted with a frictionless piston. The piston is then pushed in, thereby compressing the gas. Which statement below regarding this process is consistent with the first law of thermodynamics?
(a) The internal energy of the gas will increase.
(b) The internal energy of the gas will decrease.
(c) The internal energy of the gas will not change.
(d) The internal energy of the gas may increase, decrease, or remain the same, depending on the amount of heat that the gas gains or loses.
The statement consistent with the first law of thermodynamics is the internal energy of the gas will increase. Option a is correct.
The statement consistent with the first law of thermodynamics is (a) The internal energy of the gas will increase. This is because the work done on the gas is being converted into an increase in internal energy.
Option (b) (c) The internal energy of the gas will decrease, or will not change is incorrect, because work is being done on the gas which will increase its internal energy.
Option (d) does not address the fact that work is being done on the gas, which is also contributing to changes in its internal energy.
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the origin: 8. A thin rod of length and uniform charge per unit length SA lies along the x axis as shown in Figure P23.8. (a) Show that the electric field at P, a distance d from the rod along its perpendicular bisector, has no x component and is given by E = 2k i sin 0,/d. (b) What If? Using your result to part (a), show that the field of a rod of infinite length is E = 2kX/d. -- -- -- --- F
For solution of a) of the particular question where a thin rod and uniform change lies along the x axis=
λ = linear charge density
Consider a small length "dx" at distance "x" from the origin
dq = small charge on the small length = λ dx
r = distance of small length from point P = sqrt(x2 + d2)
small electric field at P due to small length is given as
dE = k dq/r2
dE = k λ dx /(sqrt(x2 + d2))2
dE = k λ dx /(x2 + d2)
From the diagram , we see that "dE Sinθ" are equal and opposite, hence x-components cancel out.
Net electric field at P is given as
E = ∫ 2 dE Cosθ
E = ∫ 2 (k λ dx /(x2 + d2)) (d/sqrt(x2 + d2))
E = ∫ 2 (k λ d dx /(x2 + d2)3/2)
E = ∫ _{0}^{l/2} 2 (k λ d dx /(x2 + d2)3/2)
E = (2 k λ d) ∫_{0}^{l/2}dx /(x2 + d2)3/2
E = (2 k λ d) ((l/2)/ (d2 sqrt(d2 + (l/2)2))
E = (2 k λ d) (Sinθ _{o}/ d2 )
Since
Sinθ _{o} = (l/2) /sqrt(d2 + (l/2)2)
E = 2 k λ Sinθ _{o}/ d
For solution of b)
for infinite length , θ _{o}= 90
E = 2 k λ Sin90/ d
E = 2 k λ / d
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The block then moves up a hill that is not frictionless. Determine what height the block reaches if 430 J of thermal energy is produced. On a frictionless horizontal surface, a 9.61 kg block is pushed up against a 34,596 N/m spring and compresses it 0.23 m. The block is then released. A.) Determine the block's speed after it leaves the spring. V = 13.8 m/s B.) The block then moves up a hill that is not frictionless. Determine what height the block reaches if 430 J of thermal energy is produced. h =
Answer:
51
Explanation:
you will settle down and calculate it well
the motion of the molecules reflect the kinetic enregy of molecules or is ordered and predictable reflects the potential energy of moecules and is random and erratic
The motion of molecules is best described as being random and erratic. Option d is correct answer.
The movement of molecules is a reflection of the kinetic energy they possess. The molecules in a substance are always moving, even in solid objects, but the motion is typically less than in liquids or gases. The kinetic energy of a molecule is related to its speed and mass. The faster and heavier the molecule, the more kinetic energy it has.
The motion of molecules is not ordered or predictable, meaning that they do not follow a specific path or pattern. Instead, the molecules move randomly, colliding with one another and bouncing off surfaces. This random motion is due to the thermal energy that is present in all objects. Thermal energy is the energy that causes objects to become hotter, and it is related to the potential energy of molecules.
--The given question is incomplete, the complete question is
"The motion of the molecules reflect
a. the kinetic energy of molecules
b. is ordered and predictable
c. reflects the potential energy of molecules
d. is random and erratic" --
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why is the world so small compared to the sun and jupider
The Sun appears smaller than the Earth from here on Earth, but that is only because the Earth is considerably closer to you than the Sun is. Jupiter due to its rapid revolution, which increases its diameter in the midsection.
What are the bodies of the solar system?Our solar system consists of the star, Sun, planets: Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, and Neptune, and small planets such as Pluto.
While the Sun is 150 million kilometers away from where you are, you are on the surface of the Earth.
The planet is an oblate spheroid due to its rapid revolution, which increases its diameter in the midsection.
Therefore, Jupiter due to its rapid revolution increases its diameter in the midsection making it bigger as compared to the world, so it is believed small compared to the Sun and Jupiter.
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ntroduction The flow of geophysical fluids (i.e., the Earth’s ocean and atmosphere, and the atmospheres of gas giants planets such as Jupiter and Saturn) is complicated, involving a vast number of processes and interactions among them on scales ranging from centimeters to the planet’s size, and timescales going from seconds to millennia. Two effects mainly constrain the flow of geophysical fluids: the planet’s rotation and stratification. In this lab we will deal with the first of the aforementioned effects. We will learn how the unusual properties of rotating fluids manifest themselves in, and profoundly influence, the circulation of the Earth’s ocean and planetary atmospheres. The planet’s rotation makes these fluids more similar than one might expect. On what scales might the atmosphere, ocean, or our laboratory experiment, “feel” the effect of rotation? Suppose that U is a typical horizontal current speed, and the typical distance over which the currents varies is L. Then the timescale of the motion (Tmotion) is L/U. Compare this with the period of rotation Trot, define a nondimensional number (the Rossby number):
Ro := Trot/Tmotion = Trot × U/L. If Ro is much greater than one, then the timescale of motion is short relative to a rotation period, and rotation will not significantly influence the motion. If Ro is much less than one, then the motion will be aware of rotation. Let us estimate Ro for large-scale flow in the atmosphere and ocean.
• Amosphere: L ∼ 5000 Km, U ∼ 10 m/s, and T = 1 day, giving Ro = 0.2, which suggest the rotation will be important.
• Ocean: L ∼ 1000 Km, U ∼ 0.1 m/s, giving Ro = 0.01, and rotation will be a controlling factor. Pre Lab 1.
It is clear from the Ro estimations above, that rotation is very important in shaping the patterns of air and ocean currents on sufficiently large scales. How can we study this effect on an small rotating tank (L ∼ 30cm)? If we generate a current in the tank of U ∼ 0.1 cm/s, what would be an appropriate rotation period?
Answer:
Explanation:
To study the effect of rotation on a small rotating tank, we want the Rossby number to be much less than one, so that rotation will be a controlling factor in shaping the patterns of flow.
Based on the given information, the length scale of the tank is L = 30 cm and the current speed is U = 0.1 cm/s. To calculate the timescale of the motion, Tmotion, we can use the formula Tmotion = L/U, which gives us:
Tmotion = 30 cm / 0.1 cm/s = 300 s
Next, we need to estimate an appropriate rotation period, Trot, so that the Rossby number Ro = Trot / Tmotion will be much less than one. We can use the formula Ro = Trot * U / L, rearrange it to solve for Trot:
Trot = Ro * L / U
If we take Ro to be 0.1 (for example), then we have:
Trot = 0.1 * 30 cm / 0.1 cm/s = 30 s
So, with a rotation period of 30 s and a current speed of 0.1 cm/s, we should expect the rotation to have a significant influence on the patterns of flow in the small rotating tank.
A 9.0-kg iron ball is dropped onto a pavement from a height of 140 m. Suppose that half of the heat generated goes into warming the ball.
What is the temperature increase of the ball.
(The specific heat capacity of iron is 450 J/k ⋅ ∘C. Use 9.8 N/kg for g.). Express your answer to two significant figures and include the appropriate units.
If half of the heat generated goes into warming the ball then the
temperature increase of the ball into 0.71°C
Conservation of energy: ball is dropped (Vinitial=0, KE=0) so Energy at top is PE=mgh.
This will be the same amount used to generate heat.
1/2E=1/2 mgh=1/215 kg 9.8 m/s²
m =.5159.8J
H=1/2E=cmT
.5159.8J=450 J/kgC 15kg T
T=.5159.8J / (450 J/kgoC 15kg) =.5159.8/(450*15)°C
potential energy U = m*g*h
heat = U/2
m*c * delta _T = m*g*h/2
delta_T = m*g*h/2*m*c
delta_T = g*h / 2C
delta_T = 9.8*150 / 2*450
delta_T = 1.63 degrees
What is energy?
The ability to perform work or generate heat is referred to as energy, which is aa fundamental physical characteristic. It has magnitude but no direction because it is a scalar quantity. Kinetic energy, potential energy, thermal energy, electromagnetic energy, chemical energy, and other kinds of energy can all exist.Energy cannot be created or destroyed; it can only be changed from one form to another, according to the rule of conservation of energy. This implies that the overall level of energy in a closed system doesn't change.In many disciplines, including physics, engineering, chemistry, and biology, the concept of energy is fundamental.To know more about energy, click the link given below:
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consider a particle of mass m decaying into two bodies of masses m1 and m2. find expressions for the energies of the decay products in the cm frame in terms of the masses: m, m1 and m2. find expressions for the momenta of the decay products in the cm frame in terms of the cm energies and the masses m1 and m2.
In the centre of- mass (CM) frame, the energies of the two decay products are:
[tex]$$ E_1 = \frac{(m-m_2)^2 - p^2}{2m_1}c^2 $$[/tex]
[tex]$$ E_2 = \frac{(m-m_1)^2 - p^2}{2m_2}c^2 $$[/tex]
The momenta of the two decay products are:
[tex]$$ p_1 = \frac{\sqrt{(m^4 - 2m^2(m_1^2+m_2^2) + (m_1^2-m_2^2)^2)}}{2m c} $$[/tex]
[tex]$$ p_2 = -p_1 $$[/tex]
What does Centre of-mass mean?The centre of mass (CM) is a point that represents the average position of mass in a system. In a system of particles, the CM is the point where the weighted average position of all the particles is located. It is a useful concept in physics and engineering because it allows us to simplify calculations of the motion and interactions of the system as a whole.
In the context of particle physics, the CM frame is a reference frame in which the total momentum of a system of particles is zero. This means that the particles are moving with equal and opposite momenta in this frame, and it simplifies the analysis of the system, allowing us to study its properties and interactions. The CM frame is often used in particle accelerators, where high-energy collisions between particles can produce a large number of new particles that move in various directions.
Let the initial particle of mass [tex]$m$[/tex] be at rest in the CM frame. After decay, the two particles will move in opposite directions, each with momentum [tex]$p$[/tex]The total energy of the system is conserved, and it is given by the sum of the energies of the two particles:
[tex]$$ E = E_1 + E_2 $$[/tex]
where[tex]$E_1$[/tex]and [tex]$E_2$[/tex]the energies of the two particles.
The total energy [tex]$E$[/tex] of the system is given by:
[tex]$$ E = \sqrt{(mc^2)^2 + (pc)^2} $$[/tex]
where [tex]$c$[/tex] is the speed of light.
Since the particles are moving in opposite directions, their momenta are equal in magnitude but opposite in direction, i.e., [tex]$p_1 = -p_2 = p$[/tex]. The energies of the particles can be found using the following expression:
[tex]$$ E_i = \sqrt{(m_ic^2)^2 + (p_ic)^2} $$[/tex]
where [tex]$i$[/tex] is the particle index.
Substituting[tex]$p_1 = -p_2 = p$ and $E = E_1 + E_2$[/tex] in the above equations, we get:
[tex]$$ E_1 = \frac{m_1^2c^4 + p^2c^2}{2m_1c^2} $$[/tex]
[tex]$$ E_2 = \frac{m_2^2c^4 + p^2c^2}{2m_2c^2} $$[/tex]
Solving for [tex]$p$[/tex]
[tex]$$ p = \frac{\sqrt{(E^2 - (m_1c^2 + m_2c^2)^2)(E^2 - (m_1c^2 - m_2c^2)^2)}}{2Ec} $$[/tex]
The momenta of the two particles in the CM frame are given by:
[tex]$$ p_1 = \frac{\sqrt{(E^2 - (m_1c^2 + m_2c^2)^2)}}{2c} $$[/tex]
[tex]$$ p_2 = \frac{\sqrt{(E^2 - (m_1c^2 - m_2c^2)^2)}}{2c} $$[/tex]
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experimental resistance is calculated using measured values. calculated resistance is determined using equations from your textbook or the background information link. question 1) how is the total resistance related to the individual resistances? question 2) total current to the individual currents? question 3) total voltage to the individual voltages? be sure to show your calculations for the series circuit.
Answer:
Explanation:
edge 2023 b
A hot dog cooker heats hot dogs by connecting them to 120 V household electricity. A typical hot dog has a mass of 70 g and a resistance of 160 Ω.
Part A
How long will it take for the cooker to raise the temperature of the hot dog from 20∘C to 85 ∘C? The specific heat of a hot dog is approximately 2500 J/kg⋅K
It will take about 233.34 seconds for the hot dog cooker to raise the temperature of the hot dog from 20∘C to 85∘C.
What is specific heat?Specific heat is the amount of heat energy required to raise the temperature of a substance by one degree Celsius or one Kelvin per unit mass.
Here,
To solve this problem, we can use the formula for the amount of heat required to change the temperature of an object,
Q = mcΔT
First, we need to calculate the amount of heat required to raise the temperature of the hot dog from 20∘C to 85∘C:
Q = (0.07 kg)(2500 J/kg⋅K)(85∘C - 20∘C)
Q = 1058.5 J
Next, we can use the formula for electrical power,
P = IV
We can rearrange this formula to solve for the current:
I = P/V
The power required to heat the hot dog can be calculated using the formula for electrical power:
P = V²/R
Substituting the given values, we get:
P = (120 V)²/160 Ω
P = 90 W
I = 90 W/120 V
I = 0.75 A
Finally, we can use the formula for the amount of time required to transfer a certain amount of heat:
t = Q/(IΔT)
Substituting the values we calculated, we get:
t = 1058.5 J/(0.75 A)(65∘C)
t = 233.34 s
Therefore, it will take about 233.34 seconds for the hot dog cooker to raise the temperature of the hot dog from 20∘C to 85∘C.
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problem 11.143 a race car enters the circular portion of a track that has a radius of 70 m. when the car enters the curve at point p, it is travelling with a speed of 120 km/h that is increasing at 5 m/s2 . three seconds later, determine (a) the total acceleration of the car in xy components, (b) the linear velocity of the car in xy components.
(a) The total acceleration of the car in xy components is 16.73 m/s^2.
(b) The linear velocity of the car in xy components is 33.3 m/s.
What is total acceleration of the car?To find the total acceleration of the car in xy components, we need to find the sum of the centripetal and tangential accelerations.
The centripetal acceleration is given by a = v^2/r
where;
v is the velocity of the car and r is the radius of the circular trackAt point P, the velocity of the car is 120 km/h = 33.33 m/s. Therefore, the centripetal acceleration is:
a_c = v^2/r
a_c = (33.33 m/s)^2 / 70 m
a_c = 15.88 m/s^2
The tangential acceleration is given by a_t = dv/dt
where;
v is the velocity of the car and
t is time
The rate of change of velocity is given as 5 m/s^2. Therefore, the tangential acceleration is:
a_t = 5 m/s^2
The total acceleration of the car in xy components is the vector sum of a_c and a_t. Since the two accelerations are perpendicular, we can use the Pythagorean theorem to find the magnitude of the total acceleration:
a_total = √(a_c^2 + a_t^2)
a = √((15.88 m/s^2)^2 + (5 m/s^2)^2)
a = 16.73 m/s^2
(b) The linear velocity of the car in xy components can be found using the formula;
v = rω
where;
r is the radius of the circular track and ω is the angular velocity of the car.The angular velocity is related to the linear velocity by the formula ω = v/r. At point P, the linear velocity of the car is 33.33 m/s. Therefore, the angular velocity is:
ω = v/r = 33.33 m/s / 70 m
ω = 0.476 rad/s
The linear velocity in the x-direction is v_x = v cos(θ),
where;
θ is the angle between the velocity vector and the x-axis.Since the car is entering the circular portion of the track, the angle θ is 0. Therefore, v_x = v cos(0) = 33.33 m/s.
The linear velocity in the y-direction is v_y = v sin(θ). Since the car is entering the circular portion of the track, the angle θ is 90 degrees. Therefore, v_y = v sin(90°) = 33.33 m/s.
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