Answer:
(a) α = - 1000 rad/s²
Negative sign represents deceleration.
(b) θ = 3581 rotations
(c) L = 1800 m
(d) a = - 80 m/s²
Explanation:
(a)
using First equation of motion for angular motion:
ωf = ωi + αt
where,
ωf = Final Angular Speed = 2000 rad/s
ωi = Initial Angular Speed = 7000 rad/s
α = Angular Acceleration = ?
t = time = 5 s
Therefore,
2000 rad/s = 7000 rad/s + α(5s)
α = (2000 rad/s - 7000 rad/s)/5 s
α = - 1000 rad/s²
Negative sign represents deceleration.
(b)
Using second equation of motion:
θ = ωi t + (1/2)αt²
where,
θ = No. of Rotations = ?
Therefore,
θ = (7000 rad/s)(5 s) + (1/2)(- 1000 rad/s²)(5 s)²
θ = 35000 rad - 12500 rad
θ = (22500 rad)(1 rotation/2π rad)
θ = 3581 rotations
(c)
Length of String = L = (Circumference of Pulley)(θ)
L = [2π(0.08 m)][3581 rotations]
L = 1800 m
(d)
Tangential Acceleration = a = rα
a = (0.08 m)(-1000 rad/s²)
a = - 80 m/s²
Emmy kicks a soccer ball up at an angle of 45° over a level field. She watches the ball's trajectory and notices that it lands, two seconds after being kicked, about 20 m away to the north. Assume that air resistance is negligible, and plot the horizontal and vertical components of the ball's velocity as a function of time. Consider only the time that the ball is in the air, after being kicked but before landing. Take "north" and "up" as the positive x ‑ and y ‑directions, respectively, and use g≈10g≈10 m/s2 for the downward acceleration due to gravity
Let [tex]u[/tex] be the initial velocity of the soccer ball at an angle of inclination of [tex]\theta_0[/tex] with the positive x-axis.
Given that:
[tex]\theta_0=45^{\circ}[/tex]
The horizontal distance covered by the projectile=20 m
Time of flight, [tex]t_f=2[/tex] seconds
Acceleration due to gravity, [tex]g= 10 m/s^2[/tex] downward.
As "north" and "up" as the positive x ‑ and y ‑directions, respectively.
So, [tex]g= -10 m/s^2[/tex]
As the acceleration due to gravity is in the vertical direction, so the horizontal component of the initial velocity remains unchanged.
The x-component of the initial velocity, [tex]u_x=u\cos\theta_0[/tex].
The horizontal distance covered by the projectile [tex]= u_x\times t_f[/tex]
[tex]\Rightarrow u_x\times t_f=20[/tex]
[tex]\Rightarrow u_x\times 2=20[/tex]
[tex]\Rightarrow u_x=10[/tex] m/s
So, the horizontal component of the velocity is 10 m/s which is constant and the graph has been shown in the figure (i).
Now, [tex]u\cos(45^{\circ})=10[/tex] [as [tex]u_x=u\cos\theta_0[/tex]]
[tex]\Rightarrow u=10\sqrt{2}[/tex] m/s.
The vertical component of the initial velocity,
[tex]u_y= u\sin\theta_0[/tex]
[tex]\Rightarrow u_y=10\sqrt{2}\sin(45^{\circ})[/tex]
[tex]\Rightarrow u_y=10[/tex] m/s
Let v be the vertical component of the velocity at any time instant t.
From the equation of motion,
[tex]v=u+at[/tex]
where u: initial velocity, v: final velocity, a: constant acceleration, and t: time taken to change the velocity from u to v.
In this case, we have [tex]u=u_y, a= -10 m/s^2[/tex].
So at any time instant, t.
[tex]v=u_y+(-10)t[/tex]
[tex]\Rightarrow v=10-10t[/tex]
The vertical component of the velocity, v, is the function of time and related as [tex]v=10-10t[/tex].
This is a linear equation.
At 2 second, the vertical component of the velocity
v=10-10x2=-10 m/s.
The graph has been shown in figure (ii).
It takes a ball 30 seconds to travel from point A to point B. What is the ball’s speed?
(A + B) / 30 s
(B – A) / 30 s
30 s / (A - B)
30 s / (B – A)
Answer:
Direction from A to B divided by time(30s)
Explanation:
brief description about construction of mercury thermometer with figure
A car had a constant acceleration of 6m/s? north for 2 seconds. What was thecar's change in velocity if it traveled in a straight line?
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Answer:
Dang bro what school you in?
Light travelling in air striked a flat piece of uniformly thick glass at an incident of 60. If the index refraction of the glass is 1.50 what is the angle of refraction in the glass
Answer:
30.81°
Explanation:
Angle of refraction= sin⁻¹ 0.5122 = 30.81°.
Hope this helped!
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3 Physics Questions Please Help
1. Which will hit the ground FASTER (greater ), a pencil that is dropped from the top of the table ( = 0) or a pencil that rolls of the edge of the table ( > 0)? (Assume the table is level and both pencils leave the table at the same instant.)
a. The pencil that rolls off the table will hit faster.
b. The pencil that is dropped will hit faster.
c. Both will hit at the same speed.
d. Can’t tell without knowing how fast the second pencil was rolling.
2. You are throwing a water balloon as hard as you can off the top of the bleachers. You want the balloon to hit the ground with as great a velocity as possible. Which of these actions will help the hit the ground fastest—that is, with greatest ?
a. Throwing the balloon UP at any angle of θ with an initial velocity vmax.
b. Throwing the balloon DOWN at the opposite angle (-θ) with the same initial velocity vmax.
c. Both of these options—choices (a) and (b)—will yield the same .
d. Cannot compare these options without knowing angle θ.
3. Batman, hanging from the top of a low building, sees a criminal a long-distance away pointing a gun over his head at him. (The gun has no sighting tool, and the criminal does not know physics.) Assume Batman reacts just as the criminal fires the gun. (Which of the following should Batman NOT do? Circle any/all BAD MOVES.
a. Stay where he is and hope the criminal will miss.
b. Retract his grapple and hope to dodge the bullet by quickly shooting upward.
c. Let go of the building and hope to dodge the bullet by falling downward.
d. Push off the building and hope to dodge the bullet by moving sideways.
Check: did you see the word NOT in the question?
Answer:
Number one is C.Both
Number two is A. throw upwards
Number three is B. react his grapple... shooting upward
The mass of a fully loaded Boeing 747 is about 4,082,331.33 kg. If it is cruising eastward at a
velocity of 253 m/s, what is its momentum?
A locomotive engine pulls a train with a constant force comment on whether its acceleration will increase or decrease if more coaches are added to the train
Explanation:
the acceleration will be unchanged according to newton second law of motion
A locomotive engine pulls a train with a constant force then its acceleration will increase if more coaches are added to the train because the mass of the system would increase
What is Newton's second law?Newton's Second Law states that The resultant force acting on an object is proportional to the rate of change of momentum.
Force = mass ×acceleration
As given in the problem if a locomotive engine pulls a train with a constant force
As given the force is constant but if the number of coaches increases the mass of the train increase to balance the constant force acceleration will decrease.
Thus, if a locomotive engine pulls a train with a constant force then its acceleration will increase if more coaches are added to the train because the mass of the system would increase.
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Two parallel plates having charges of equal magnitude but opposite sign are separated by 29.0 cm. Each plate has a surface charge density of 32.0 nC/m2. A proton is released from rest at the positive plate. (a) Determine the magnitude of the electric field between the plates from the charge density. kN/C (b) Determine the potential difference between the plates. V (c) Determine the kinetic energy of the proton when it reaches the negative plate. J (d) Determine the speed of the proton just before it strikes the negative plate. km/s (e) Determine the acceleration of the proton. Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. m/s2 towards the negative plate (f) Determine the force on the proton. N towards the negative plate (g) From the force, find the magnitude of the electric field. kN/C (h) How does your value of the electric field compare with that found in part (a)
Answer:
(a) E = 3.6 x 10³ N/C = 3.6 KN/C
(b) ΔV = 1044 Volts
(c) K.E = 1.67 x 10⁻¹⁶ J
(d) Vf = 4.47 x 10⁵ m/s
(e) a = 3.45 x 10¹¹ m/s²
(f) F = 5.76 x 10⁻¹⁶ N
(g) E = 3.6 x 10³ N/C = 3.6 KN/C
(h) Both values are same in part (h) and (a)
Explanation:
(a)
Electric field between oppositely charged plates is given as follows:
E = σ/ε₀
where,
E = Electric Field Intensity = ?
σ = surface charge density = 32 nC/m² = 3.2 x 10⁻⁸ C/m²
ε₀ = Permittivity of free space = 8.85 x 10⁻¹² C²/N.m²
Therefore,
E = (3.2 x 10⁻⁸ C/m²)/(8.85 x 10⁻¹² C²/N.m²)
E = 3.6 x 10³ N/C = 3.6 KN/C
(b)
E = ΔV/r
ΔV = Er
where,
r = distance between plates = 29 cm = 0.29 m
ΔV = Potential Difference = ?
ΔV = (3.6 x 10³)(0.29)
V = 1044 Volts
(c)
Kinetic Energy of Proton = Work done on Proton
K.E = F r
but, F = E q
K.E = E q r
where,
q = charge on proton = 1.6 x 10⁻¹⁹ C
Therefore,
K.E = (3600 N/C)(1.6 x 10⁻¹⁹ C)(0.29 m)
K.E = 1.67 x 10⁻¹⁶ J
(d)
K.E = (1/2)m(Vf² - Vi²)
where,
m = mass of proton = 1.67 x 10⁻²⁷ kg
Vf = Final Velocity = ?
Vi = Initial Velocity = 0 m/s
Therefore,
1.67 x 10⁻¹⁶ J = (1/2)(1.67 x 10⁻²⁷ kg)(Vf² - (0 m/s)²]
Vf² = (1.67 x 10⁻¹⁶ J)(2)/(1.67 x 10⁻²⁷ kg)
Vf = √(20 x 10¹⁰ m²/s²)
Vf = 4.47 x 10⁵ m/s
(e)
2as = Vf² - Vi²
2(a)(0.29 m) = (4.47 x 10⁵ m/s)² - (0 m/s)²
a = (20 x 10¹⁰ m²/s²)/0.58 m
a = 3.45 x 10¹¹ m/s²
(f)
F = ma
F = (1.67 x 10⁻²⁷ kg)(3.45 x 10¹¹ m/s²)
F = 5.76 x 10⁻¹⁶ N
(g)
E = F/q
E = (5.76 x 10⁻¹⁶ N)/(1.6 x 10⁻¹⁹ C)
E = 3.6 x 10³ N/C = 3.6 KN/C
(h)
Both values are same in part (h) and (a)
if a 100-n net force acts upon a 50g car, what will the acceleration of the car be
Answer:
2m/s
Explanation:
The formula for acceleration can be found by [ a = f/m ]
We are given the force [ 100N ] and the mass [ 50g ].
We can use these values to solve for the acceleration.
a = 100/50
a = 2m/s
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Whenever I visit one of the big membership-driven bulk wholesalers in town, without fail I end up using a junk cart with bum wheels that not only drag, but pull to the left. If a cart requires 20 lb of force forward to overcome the wheel’s drag, but there are also around 3 lb of force pulling it to the left, with what force and in what direction do I need to push in order to make the cart go forward?
Required:
Measure your angle from the forward direction and specify whether it’s to the left or right.
Answer: Force of magnitude 20.22lbf to the right at angle 8.5°
Explanation: The image named "Untitled" below shows the forces acting on the cart.
The second image, "Untitled2", demonstrate the vector diagram of the forces acting on the kart, in which:
f is force forward
f(left) is pull to the left
f(R) is the resulting force necessary for the kart to move
The drag force is not draw on the vector diagram because its value is already part of the force forward.
As we can see, the diagram forms a right triangle with f(R) as hypotenuse.
Using Pytagorean Theorem:
[tex][f(R)]^{2}=f^{2}+[f(left)]^{2}[/tex]
[tex][f(R)]^{2}=20^{2}+3^{2}[/tex]
[tex][f(R)]^{2}=409[/tex]
[tex]f(R)=\sqrt{409}[/tex]
f(R) = 20.22
Force forward equals 20.22lbf to the right.
The angle formed form the forward direction and the resulting force is
[tex]cos\theta=\frac{20}{20.22}[/tex]
cosθ = 0.989
θ = cos⁻¹(0.989)
θ = 8.5°
The forward force is to the right at angle 8.5° and magnitude 20.22lbf.
Equipotential lines are lines with equal electric potential (for example, all the points with an electric potential of 5.0 V). Using the plot tool that comes with voltmeter (pencil icon) make two equipotential lines at r=0.5 m and r=1.5 m. Enable electric field vectors in the simulation. Put an electric field sensor at different points on the equipotential line and note the direction of the electric field vector. What can you conclude about the direction of the electric field vector in relation to the equipotential lines?Equipotential lines are lines with equal electric potential (for example, all the points with an electric potential of 5.0 V). Using the plot tool that comes with voltmeter (pencil icon) make two equipotential lines at r=0.5 m and r=1.5 m. Enable electric field vectors in the simulation. Put an electric field sensor at different points on the equipotential line and note the direction of the electric field vector. What can you conclude about the direction of the electric field vector in relation to the equipotential lines?
Answer:
the equipotential lines or surface are perpendicular to the electric field
Explanation:
In this exercise you are asked to explain the relationship between the direction of the equipotential lines and the electric field.
Electric field lines go from a positive charge to a negative charge, the specific shape of these lines depends on the charge distribution, let's look at two specific cases:
* in the case of point charge the lines are radial
* In the case of electric plates, they are lines perpendicular to the plates.
Equipotential surfaces are related to the elective field by
Ex = - dV / dx
Ey = -dV / dy
Ez = -dV / dz
E = Exi ^ + Ey j ^ + Ezk ^
remember electric potential is a scalar and the electric field is a vector quantity, therefore the equipotential lines or surface are perpendicular to the electric field. Specifically for
* point charge the equipotential surfaces are concentric with the charge
* For electrical plates equipotential lines are parallel to the plates.
How much work is done when 25 N of force is used to push a crate 3.0 m
Answer:
The answer is 75 JExplanation:
The work done by an object can be found by using the formula
work done = force × distanceFrom the question
force = 25 N
distance = 3 m
We have
work done = 25 × 3
We have the final answer as
75 JHope this helps you
EM waves are used for communication with satellites and for observing the Earth from space. True or false
Answer:
true
Explanation:
Order the circuits according to their power usage, from highest to lowest. All batteries are the same voltage, and all light bulbs have the same resistance. A battery connected to three light bulbs. The positive battery terminal connects to the first bulb. The first bulb connects to the second bulb. The second bulb connects to the third bulb. The third bulb connects to the negative battery terminal. Circuit A A battery connected to two light bulbs. The positive and negative terminals of the battery are connected directly to each bulb. Circuit B A battery connected to two light bulbs. The positive battery terminal connects to the first bulb. The first bulb connects to the second bulb. The second bulb connects to the negative battery terminal. Circuit C A battery connected to three light bulbs. The positive and negative terminals of the battery are connected directly to each bulb. Circuit D Highest power usage Lowest power usage
Answer:
P_a = 3P₀ , P_b = 2 P₀ , P_c = 3 P₀
P_c> P_b> P_a
Explanation:
To determine which circuit consumes more energy than is given by the expression
P = V I
V = I R
I = V / R
P = V² / R
As all circuits have the same battery, the value of the resistance to which the battery is connected determines the consumption
Circuit A
In this circuit the three bulbs are in series so the total resistance is
R_total = 3 R
the power dissipated is
P_a = V² / 3R
if we call
P₀ = V² / R
we substitute
P_a = P₀/3
Circuit B
Two bulbs are connected in series
R_total = 2 R
power is
P_b = V2 / 2R
P_b = P₀/2
Circuit C
The 3 bulbs are connected, but in parallel, the resistance is
1 / R_totak = 1 / R + 1 / R + 1 / R
R_total = R / 3
P _c = V2 3 / R
P_c = 3 Po
By reviewing these results, we can sort the circuits
P_c> P_b> P_a
A 90kg woman and a 60kg boy are standing at rest on a frictionless frozen lake. The boy pushes the woman with a 40N horizontal force. Calculate the acceleration of the woman
3600 m/s2
0.66 m/s2
26.7 m/s2
0.44 m/s2
30 m/s2
1.5 m/s2
What happens to them?
The woman will move slowly, and the boy will move faster in the opposite direction.
The woman and the boy will move together with equal speeds.
The woman will move away, but the boy will remain at rest.
The boy will move away, but the woman will remain at rest.
The woman will move slowly, and the boy will move faster in the same direction.
The woman and the boy will move apart at equal speeds.
Answer:
The acceleration of the woman is 0.44 m/s²
Explanation:
Given;
mass of the woman, m₁ = 90 kg
mass of the boy, m₂ = 60 kg
The force applied by the boy, f₂ = 40 N
The net horizontal force on the woman = 40 N
Apply Newton's second law of motion to determine the acceleration of the woman;
f = ma
a = f / m
a = 40 / 90
a = 0.44 m/s²
Therefore, the acceleration of the woman is 0.44 m/s²
Why are some substances dissolve both polar and nonpolar substances?
Answer:
Because water is polar and oil is nonpolar, their molecules are not attracted to each other. polar solvents dissolve polar solutes, and nonpolar solvents dissolve nonpolar solutes.
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Substances that can dissolve both polar and nonpolar substances are called amphiphilic or amphipathic. This unique property arises due to their molecular structure, which contains both polar and nonpolar regions.
Amphiphilic molecules typically have a hydrophilic (water-attracting) or polar end and a hydrophobic (water-repelling) or nonpolar end. This dual nature allows them to interact with and dissolve in both polar and nonpolar solvents. Let's explore the reasons behind this ability:
Polar interactions: The hydrophilic end of the amphiphilic molecule can form hydrogen bonds and electrostatic interactions with polar solvents (like water) due to the presence of charged or partially charged atoms (such as oxygen or nitrogen). This makes them soluble in polar substances.
Nonpolar interactions: The hydrophobic end of the amphiphilic molecule is typically a long hydrocarbon chain or a nonpolar group. These nonpolar regions can interact with other nonpolar substances (like oils, fats, or hydrophobic portions of biomolecules) through van der Waals forces and London dispersion forces.
Emulsification: The presence of both polar and nonpolar parts in an amphiphilic molecule allows it to act as an emulsifier. It can stabilize the interface between polar and nonpolar substances, preventing them from separating. For example, in the formation of an oil-in-water or water-in-oil emulsion, amphiphilic molecules form a barrier between the two immiscible substances, keeping them dispersed and forming a stable mixture.
Examples of amphiphilic molecules include surfactants (detergents), phospholipids (essential components of cell membranes), and lipoproteins (transporters of lipids in the bloodstream). These substances play vital roles in various biological processes and industrial applications due to their ability to interact with both polar and nonpolar substances.
Hence, Substances that can dissolve both polar and nonpolar substances are called amphiphilic or amphipathic. This unique property arises due to their molecular structure, which contains both polar and nonpolar regions.
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To increase the current in a circuit, which can be increased?
voltage
resistance
interference
ohms
Answer:
voltage
Explanation:
the current in a circuit, the voltage can be increased.
What is ohm's law?According to Ohm's law, when all other physical parameters, including temperature, are held constant, the voltage across a conductor is directly proportional to the current flowing through it.
From ohm's law, we can state that voltage is directly proportional to current.
Looking at the following formula:
I = V/R
Where V is voltage,
I is current, and
R is resistance
In the above equation, Current and Voltage are in a direct relationship such that if I is increased, V is also increased, and vice versa.
Therefore, To increase the current, the voltage should be increased.
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Which statement(s) is/are TRUE about photons?
SELECT ALL THAT APPLY
a Photons have greater mass than electrons.
b Photons are particles of light.
с
Photons carry a very specific quantum of
energy.
d Photons have zero mass.
Answer:
Not D, every substance in existence has mass and is made from it in any given form.
B and C
Explanation:
The statements which are true about photons are:
Choice B: Photons are particles of light.Choice C: Photons carry a very specific quantum of energy.Choice D: Photons have zero mass.Discussion:
The photon is a type of elementary particle. It is the quantum of the electromagnetic field including electromagnetic radiation such as light and radio waves, and the force carrier for the electromagnetic force.
Additionally, Photons are massless, so they always move at the speed of light in vacuum, 3 × 10⁸ m/s.
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A volleyball player jumps to hit a ball horizontally at 7.0 m/s straight on. If the
height at which the ball was hit is 3.0 m tall, how far did the ball go horizontally
before it hit the ground?
5.5 m
3.6 m
O 4.3 m
4.2 m
Answer:
5.5 is the correct answer
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The moon is a projectile that ...
A. Is big enough to keep its momentum around the earth
B. Is far enough to avoid the gravity of the earth
C. Has enough velocity to fall around the earth
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Calculate Gravitational Potential Energy for an object on Earth with a mass of 2 kg and a height of 7 m.
Answer:
137.2J
Explanation:
Ep= mgh
Given,
m=2kgh=7mand we know, g = 9.81 N/kg
Ep= 2 × 9.81 × 7
Ep= 137.2J
How do astronomers determine the surface temperature of stars? *
Answer:
To the extent that Stellar spectra look like blackbodies, the temperature of a star can also be measured amazingly accurately by recording the brightness in two different filters. To get a stellar temperature: Measure the brightness of a star through two filters and compare the ratio of red to blue light.
Explanation:
Look at the image if you need me to add the answers so you can see them better i will
Answer:
0.75 meters per second; north
Explanation:
speed (velocity)= distance/time
15 meters= distance
20 seconds= time
15/20= 0.75 meters per second; north
Hope this helps!
Vector B~ has x, y, and z components of 7.1,
8.2, and 8.4 units, respectively.
Calculate the magnitude of B~ .
Can someone please help me?
Answer:
The magnitude of vector B is 13.7
Explanation:
Given that,
Component of vector B is
[tex]x=7.1[/tex]
[tex]y=8.2[/tex]
[tex]z=8.4[/tex]
We need to calculate the magnitude of vector B
Using given data
[tex]B=\sqrt{x^2+y^2+z^2}[/tex]
Put the value into the formula
[tex]B=\sqrt{(7.1)^2+(8.2)^2+(8.4)^2}[/tex]
[tex]B=13.7[/tex]
Hence, The magnitude of vector B is 13.7
If the net force acting on an object is 100 N to the left, Are the forces unbalanced or balanced?
Answer: STOP LOOKING FOR ANSWERS OMG:(((((9
Explanation:
What is the acceleration of the object?
m/s2
Answer:
nice app is brainly heheh........
Answer:
The acceleration is 100m
Explanation:
3. An 18.0 N force pulls a cart against a 15.0 N frictional force. The speed of the cart increases 1.0 m/s every 5.0 s. What is the cart's mass?
Answer:
The total force on the cart is:
F = F(applied) - F(friction) = (18 N) - (15 N) = 3.0 N
The acceleration of the cart is:
a = (change in velocity)/(time) = (1.0 m/s)/(5.0 s) = 0.20 m/s^2
Using F = ma, the mass of the cart is:
m = F/a = (3.0 N)/(0.20 m/s^2) = 15 kg
A horse slows from a velocity of 9.5 m/s to 5.5 m/s over a distance of 32 m. Find time.
Answer:
8
Explanation:
Finding time= t=change in velocity or distance/acceleration
so 32/9.5-5.5
32/4
8