Answer:
|F| = 35 kN
Explanation:
a = F/m
100 km/hr(1000 m/km / 3600 s/hr) = 27.8 m/s
v² = u² + 2as
a = (v² - u²) / 2s
F/m = (v² - u²) / 2s
F = m(v² - u²) / 2s
F = 5000(0² - 27.8²) / 2(55)
F = - 35,072.9517...
Calculate the torque produced by a 50.0 N perpendicular force at the end of a 0.300 m long wrench.
Answer:
Torque = 50N x 0.3m = 15Nm
Explanation:
Torque = Force x length of lever arm. To obtain the torque simply multiply the two given values.
Power can be defined as?
A. The distance over which work has done
B. How much work can be done in a given time
C. All the work in an given area
D. The energy required to do work
( Last question was wrong according to the test I took)
Students are asked to create roller coasters for marbles. Their goal is to design a coaster with the tallest possible hill that a marble released from a height of 1.5 m (meters) can clear. The marbles will experience some air resistance and friction as
they move.
What should the students keep in mind as they build their designs?
a)The hill can be taller than 1.5 m (meters), because the marble will be moving faster than its initial velocity allowing it to travel higher than its release height.
b)The hill can be taller than 1.5 m (meters), because the marble will gain mechanical energy as it moves allowing it to travel higher than its release height.
c)The hill should be a little less than 1.5 m (meters) high, because the marble will lose mechanical energy as it moves, preventing it from reaching its release
height.
d)The hill can be exactly 1.5 m (meters) high, because mechanical energy is always
conserved allowing the marble to travel to its release height.
Answer:
Kinetic Energy.
Explanation:
The movement of a roller coaster is accomplished by the conversion of potential energy to kinetic energy. The roller coaster cars gain potential energy as they are pulled to the top of the first hill. As the cars descend the potential energy is converted to kinetic energy.
Since the marble will loose mechanical energy, the hill should be a little less than 1.5 m (meters) high.
A roller coaster is used to demonstrate the conversion of mechanical energy. In a roller coaster, potential energy is converted to a kinetic energy hence it conveniently serves as a device for demonstrating energy conversions.
As the students make their design, they must bear in mind that the hill should be a little less than 1.5 m (meters) high, because the marble will lose mechanical energy as it moves, preventing it from reaching its release
height.
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The coefficient of kinetic friction for a 22 kg bobsled on a track is 0.10. What force is required to push it down a 5.0 degree incline and achieve a speed of 62 km/h at the end of 75 m
Hi there!
We must begin by converting km/h to m/s using dimensional analysis:
[tex]\frac{62km}{1hr} * \frac{1hr}{3600sec}*\frac{1000m}{1km} = 17.22 m/s[/tex]
Now, we can use the kinematic equation below to find the required acceleration:
vf² = vi² + 2ad
We can assume the object starts from rest, so:
vf² = 2ad
(17.22)²/(2 · 75) = a
a = 1.978 m/s²
Now, we can begin looking at forces.
For an object moving down a ramp experiencing friction and an applied force, we have the forces:
Fκ = μMgcosθ = Force due to kinetic friction
Mgsinθ = Force due to gravity
A = Applied Force
We can write out the summation. Let down the incline be positive.
ΣF = A + Mgsinθ - μMgcosθ
Or:
ma = A + Mgsinθ - μMgcosθ
We can plug in the given values:
22(1.978) = A + 22(9.8sin(5)) - 0.10(22 · 9.8cos(5))
A = 46.203 N
Students perform a set of experiments by placing a block of mass m against a spring, compressing the spring a distance x along a horizontal surface of negligible friction, releasing the block, and measuring the velocity v of the block as it leaves the spring, as shown in Figure 1. The experiments indicate that as x increases, so does v in a linear relationship. The surface is now lifted so that the surface is at an angle θ above the horizontal. Which of the following indicates how the relationship between v and x changes?
Increasing the angle of inclination of the plane decreases the velocity of the block as it leaves the spring.
The statement that indicates how the relationship between v and x changes is; As x increases, v increases, but the relationship is no longer linear and the values of v will be less for the same value of x.Reasons:
The energy given to the block by the spring = [tex]\mathbf{0.5 \cdot k \cdot x^2}[/tex]
According to the principle of conservation of energy, we have;
On a flat plane, energy given to the block = [tex]0.5 \cdot k \cdot x^2[/tex] = kinetic energy of
block = [tex]0.5 \cdot m \cdot v^2[/tex]
Therefore;
0.5·k·x² = 0.5·m·v²
Which gives;
x² ∝ v²
x ∝ v
On a plane inclined at an angle θ, we have;
The energy of the spring = [tex]\mathbf{0.5 \cdot k \cdot x^2}[/tex]
The force of the weight of the block on the string, [tex]F = m \cdot g \cdot sin(\theta)[/tex]
The energy given to the block = [tex]0.5 \cdot k \cdot x^2 - m \cdot g \cdot sin(\theta)[/tex] = The kinetic energy of block as it leaves the spring = [tex]\mathbf{0.5 \cdot m \cdot v^2}[/tex]
Which gives;
[tex]0.5 \cdot k \cdot x^2 - m \cdot g \cdot sin(\theta) = 0.5 \cdot m \cdot v^2[/tex]
Which is of the form;
a·x² - b = c·v²
a·x² + c·v² = b
Where;
a, b, and c are constants
The graph of the equation a·x² + c·v² = b is an ellipse
Therefore;
As x increases, v increases, however, the value of v obtained will be lesser than the same value of x as when the block is on a flat plane.Please find attached a drawing related to the question obtained from a similar question online
The possible question options are;
As x increases, v increases, but the relationship is no longer linear and the values of v will be less for the same value of xThe relationship is no longer linear and v will be more for the same value of xThe relationship is still linear, with lesser value of vThe relationship is still linear, with higher value of vThe relationship is still linear, but vary inversely, such that as x increases, v decreasesLearn more here:
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what are the greenhouse gasses in the earth that are primarily responsible for the greenhouse effect on Earth
Answer:
Water Vapour, Carbon dioxide, methane,nitrous oxide,ozone.
Which statement is true?
O A ball with more kinetic energy rolls farther because it is harder to stop.
A ball at its highest speed demonstrates its greatest potential energy.
When a ball comes to a stop, its kinetic energy is greatest.
A ball with less kinetic energy rolls farther because it is harder to stop.
Answer:
c when a ball stop it's kinetic enegy is the greatest
When a ball comes to a stop, its kinetic energy is greatest is true.
What are the types of energy ?The energy is the ability to do work or produce action and / or movement and manifests itself in many different ways, such as body movement, heat, electricity, etc.
The various types of energy include Kinetic energy which is associated with the movement of bodies and the Potential energy is stored by virtue of a body's position also called gravitational potential energy.
Thermal Energy can be referred as the energy associated with the kinetic energy of the molecules that make up an element, it can be manifested if there is a temperature difference between two bodies.
Chemical energy released or formed from chemical reactions, Solar energy from sunlight. This form of energy is used to generate electricity through photovoltaic plates, for example.
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An archer's bow is drawn at its midpoint until the tension in the string is 0.842 times the force exerted by the archer. What is the angle between the two halves of the string
Consult the attached free body diagram.
If we take the direction of F to be the positive horizontal axis, and upward to be the positive vertical axis, then using Newton's second law we have net forces
• ∑ F [horizontal] = F [archer] + T cos(180° - θ) + T cos(180° + θ) = 0
• ∑ F [vertical] = T sin(180° - θ) + T sin(180° + θ) = 0
since the bow is held in place while it's drawn. T is the magnitude of the tension in the string, and it can be shown to be equal in both strings since they both make the same angle with the negative horizontal axis (the dashed line).
We only really need the first equation. Simplifying it, we get
F [archer] - T cos(θ) - T cos(θ) = 0
F [archer] - 2T cos(θ) = 0
F [archer] = 2T cos(θ)
cos(θ) = F [archer] / (2T)
We're given that the tension T in the string is 0.842 times the force exerted by the archer, which is to say
T = 0.842 F [archer]
and from this we have
cos(θ) = F [archer] / (2 • 0.842 F [archer])
cos(θ) = 1/1.684
cos(θ) ≈ 0.593
Solving for θ gives an angle of θ ≈ arccos(0.593) ≈ 53.6°. Then the angle between the two tension forces is twice this, or about 2θ ≈ 107°.
A tuning fork of frequency 254 Hz and an open orang pipe of slightly lower frequency are at 15oC. When
sounded together, they produce 4 beats per second. On altering the temperature of the air in the pipes,
it is observed that the number of beats per second first diminishes to zero and then increases again to 4.
By how much has the temperature of the air in the pipe been altered?
The temperature of the air in the open orang pipe has been altered by 18.73° C
The frequency of an open orang pipe is estimated by using the formula:
[tex]\mathbf{f = \dfrac{v}{2L}}[/tex]
Then, the combination of the frequency of the tuning fork and the open orang pipe is:
[tex]\mathbf{254 - \dfrac{v}{2L} }[/tex]
These combinations of frequency produce 4 beats per sound.
i.e.
[tex]\mathbf{254 - \dfrac{v}{2L} =4}[/tex]
[tex]\mathbf{ \dfrac{v}{2L} = 254-4 }[/tex]
[tex]\mathbf{ \dfrac{v}{2L} = 250 ----(1)}[/tex]
When it is altered, the beats first diminish and increase again by 4.
i.e.
[tex]\mathbf{ \dfrac{v'}{2L} = 254+4 }[/tex]
[tex]\mathbf{ \dfrac{v'}{2L} = 258 --- (2) }[/tex]
If we equate both equations (1) and (2) together, we have:
[tex]\mathbf{\dfrac{v'}{v}= \dfrac{258}{250}}[/tex]
However, from our previous knowledge, we understand that the velocity of an object varies directly proportional to the square root of its temperature.
Hence;
when the temperature of the pipe = unknown ???the temperature of the open orang pipe = 15∴
[tex]\implies \mathbf{\sqrt{\Big(\dfrac{273 + T}{273 + 15}\Big)}= \dfrac{258}{250}}[/tex]
By squaring both sides, we have:
[tex]\implies \mathbf{\Big(\dfrac{273 + T}{273 + 15}\Big)}= \Big (\dfrac{258}{250}\Big )^2}[/tex]
[tex]\implies \mathbf{\Big(\dfrac{273 + T}{273 + 15}\Big)= \Big (\dfrac{66564}{62500}\Big )}[/tex]
[tex]\implies \mathbf{\Big(\dfrac{273 + T}{288}\Big)= \Big (1.065024\Big )}[/tex]
[tex]\implies \mathbf{273 +T =306.726912 }[/tex]
T = 306.726912 - 273
T ≅ 33.73 ° C
∴
The change in temperature ΔT = 33.73° C - 15° C
The change in temperature ΔT = 18.73° C
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9. The acceleration (a)-time (t) graph of a particle moving in a straight line is as shown in figure. At time t = 0, the velocity of particle is 10 m/s. What is the velocity at t = 8 s?
(1) 2 m/s
(2) 4 m/s
(3) 10 m/s
(4) 12 m/s
Answer:Acceleration - time graph for a particle moving in a straight line is as shown in figure. Change in velocity of the particle from t = 0 to t = 6s is:-.
1 answer
·
Top answer:
Change in velocity = (sum of area of graph) = ( 12 × 4 × 4 ) + ( 12 × ( + 2) ( - 1) ) - 4 = 8 - 4 = 4 x
Explanation:
What diameter telescope is needed to resolve the separation between an Earth-like planet and its star at 550 nm if the linear separation between them is 1 AU and the star system is 3 pc from Earth
The Rayleigh criterion allows finding the result for the diameter of the telescope that allows solving the separation of the star and the planet is:
The diameter of the telescope is D = 0.415 m
The Rayleigh criterion is used to find the separation of two points, it is based on the fact that the diffraction maximuum pattern of the first object coincides with the first minimum of the second object.
By entering in the diffraction ratio for slits you will find.
sin θ = [tex]\frac{\lambda}{a}[/tex]
In general in diffraction experiments the angles are very small,
[tex]tan \theta = \frac{y}{x} = \frac{sin \theta}{cos \theta} \\sin \theta = \frac{y}{x}[/tex]
For the case of circular apertures, when solving in polar coordinates, a constant appears.
[tex]\frac{y}{x} = 1.22 \frac{\lambda}{D}[/tex]
[tex]D = 1.22 \frac{\lambda \ x}{y}[/tex]
Where λ is the wavelength of light and D is the diameter of the aperture.
They indicate that the separation between the star and the planet is 1 AU and the distance from the system to the Earth is 3 parce.
Let's reduce the parce to astronomical units
x = 3 pc ( [tex]\frac{206264 AU}{1 pc}[/tex] )
x = 6.18 10⁵ AU
Let's calculate
D = [tex]D = 1.22 \ \frac{550 \ 10^{-9 } \ 6.18 \ 10^5 }{1}[/tex]
D = 0.415 m
In conclusion, using the Rayleigh criterion we can find the result for the diameter of the telescope that allows solving the separation of the star and the planet is:
The diameter of the telescope is D = 0.415 m
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A projectile is launched with a horizontal velocity of 20 m/s and an initial vertical velocity of 20 m/s. What is the projectile's acceleration in the Horizontal direction? Verticle direction?
Answer:
Vertical acceleration 9.8 m/s² downward
Horizontal acceleration 0.0 m/s²
assuming no air resistance.
Question below...........................
[tex]\boxed{\sf PE=mgh}[/tex]
[tex]\boxed{\sf KE=\dfrac{1}{2}mv^2}[/tex]
[tex]\boxed{\sf ME=KE+PE}[/tex]
#1
[tex]\\ \sf\longmapsto PE=60(0)(10)=0J[/tex]
[tex]\\ \sf\longmapsto KE=\dfrac{1}{2}(60)(8)^2=30(64)=1920J[/tex]
[tex]\\ \sf\longmapsto ME=1920+0=1920J[/tex]
#2
[tex]\\ \sf\longmapsto PE=60(10)(1)=600J[/tex]
[tex]\\ \sf\longmapsto KE=600J[/tex]
[tex]\\ \sf\longmapsto ME=1200J[/tex]
Now
[tex]\\ \sf\longmapsto \dfrac{1}{2}mv^2=600\implies 30v^2=600\implies v^2=20\implies v=4.2m/s[/tex]
A rod of length 0.82 m, rotating with an angular speed, 4.2 rad/s, about axes that pass perpendicularly through one end, has a mass of 0.63 kg, which is distributed uniformly along its length. Find the kinetic energies of the rod.
Hi there!
[tex]\large\boxed{KE =1.245 J}}[/tex]
The equation of angular kinetic energy is:
[tex]KE = \frac{1}{2}Iw^2[/tex]
Where:
I = moment of inertia (kgm²/s)
ω = angular speed (rad/sec)
A rod rotated about one of its endpoints has a standard moment of inertia of 1/3mR², so:
[tex]KE = \frac{1}{2}(\frac{1}{3}mL^2w^2)[/tex]
[tex]KE = \frac{1}{6}mL^2w^2[/tex]
Plug in the given values:
[tex]KE = \frac{1}{6}(0.63)(0.82^2)(4.2^2) = 1.245 J[/tex]
PLZ HURRY
When you hold a racquet and swing your arm toward the ball, there are two kinds of resistance working against your muscles—the ______ and the ______.
A.
racket, air
B.
air, ball
C.
ball, net
D.
air, racket
Answer:
It should just be A
Explanation:
I dont see the difference between these 2 but ill choose A and update you
Answer:
A: racket, air
Explanation:
when a torque is acting on a fly wheel the angular velocity of the fly wheel changes from 10rad/sec to 25rad/sec in 5sec.what will be the magnitudes of the angular acceleration of the fly wheel?
Hello!
We can use the following angular kinematic equation to solve:
α = Δω/Δt, or (ωf-ωi)/t
Plug in the given values:
α = (25 - 10)/5 = 15/5 = 3 rad/sec²
what is democratic means in science
Answer:
relating to or supporting democracy or its principles.
Explanation:
What was different about the molecules you needed to make protein 3 compared to the molecules you used to make protein 2?
Answer:
the different about the molecules we needed to make protein 3 compared to the molecules we used to make protein 2 is that if we used 2 molecules than it will be smaller than using protein 3.
QUESTION 5 When an instrument is sounded together with a turning fork of frequency 260Hz , 2 beats are heard. When the same instrument is sounded with a fork of frequency 256H2, , 6 beats are heard. Find the frequency of the instrument .
Answer:
Explanation:
4126h2.
A solenoid 91.0 cm long has a radius of 1.50 cm and a winding of 1300 turns; it carries a current of 3.60 A. Calculate the magnitude of the magnetic field inside the solenoid.
The magnitude of the magnetic field inside the solenoid is [tex]6.46 \times 10^{-3} \ T[/tex].
The given parameters;
length of the solenoid, L = 91 cm = 0.91 mradius of the solenoid, r = 1.5 cm = 0.015 mnumber of turns of the solenoid, N = 1300 current in the solenoid, I = 3.6 AThe magnitude of the magnetic field inside the solenoid is calculated as;
[tex]B = \mu_0 nI\\\\B = \mu_o(\frac{ N}{L} )I\\\\[/tex]
where;
[tex]\mu_o[/tex] is the permeability of frees space = 4π x 10⁻⁷ T.m/A
[tex]B = (4\pi \times 10^{-7}) \times (\frac{1300}{0.91} ) \times 3.6\\\\B = 6.46 \times 10^{-3} \ T[/tex]
Thus, the magnitude of the magnetic field inside the solenoid is [tex]6.46 \times 10^{-3} \ T[/tex].
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PLEASE HELP! I USED 100 POINTS!
Complete the sentence.
_____ are cracks in rock layers, and the pressures within the crust can push one of these layers past another.
A. Faults
B. Tectonic plates
C. Stalagmites
Subject: Science
Answer:
A i think its right im postivite
Answer: A. Faults
Explanation: ;-; its easy because faults are cracks and they can push layers past each other making tectonic plates rub against each other then creating a earthquake
A fat wire, radius a, carries a constant current I, uniformly distributed over its cross section. A narrow gap in the wire, of width w << a, forms a parallel-plate capacitor. Find the magnetic field in the gap, at a distance s < a from the axis.
The magnetic field in the gap is [tex]\mathbf{B = \dfrac{\mu_oI s}{2 \pi a^2}}[/tex]
In a circuit, a magnetic flux is circulated or followed via a confined environment or passage called a magnetic field gap. The narrow air gap is a non-magnetic component of a magnetic circuit that is normally connected to the remainder of the circuit magnetically in series. This enables a significant amount of magnetic flux to pass via the gap.
The magnetic field in the gap at a distance s < a can be computed by using the formula:
[tex]\mathbf{ \oint Bdl = \mu_oI_{enclosed}}[/tex]
where;
Magnetic flux density = Bdistance = d[tex]\mathbf{B( 2 \pi d) = \mu _o \oint _s J_d da }[/tex]
where;
[tex]\mathbf{J_d}[/tex] = drift current density[tex]\mathbf{B( 2 \pi d) = \mu _o J_d \oint _sda }[/tex]
[tex]\mathbf{B( 2 \pi d) = \mu _o J_d (\pi d^2) }[/tex]
Making the magnetic flux density the subject, we have:
[tex]\mathbf{B =\dfrac{ \mu _o J_d (\pi d^2) }{( 2 \pi d)}}[/tex]
[tex]\mathbf{B =\dfrac{ \mu _o J_dd}{ 2 }}[/tex]
Recall that, the drift current density [tex]\mathbf{J_d = \dfrac{I}{\pi a^2}}[/tex]
[tex]\mathbf{B = \dfrac{\mu_o d}{2}(\dfrac{I}{\pi a^2})}[/tex]
Recall that distance in question is said to be (s);
∴
[tex]\mathbf{B = \dfrac{\mu_oI s}{2 \pi a^2}}[/tex]
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A weather forecaster uses a computational model on a Monday to predict the weather on Friday. Why might that forecast change? (1 point)
A. The forecaster may have done the calculations a second time because he made a mistake in the calculation.
B. Someone may have reported the weather incorrectly before the first computation.
C. The forecaster may have had the computer model do the calculations a second time, and he found that the prediction changed.
D. An area of low pressure might move more quickly on Tuesday and Wednesday than expected.
Answer:
(d) is your answer for your question
of the following which is the largest body?
a. the moon
b. Pluto
c. Mercury
d. Ganymede
Answer:
Ganymede is the largest body
Explanation:
it is the satellite of jupiter
5 What is the maximum speed at which a car round a curve of 25m radius on a level road if the coefficient of static friction between the tires and the road is 0.80?
Hi there!
On a level road:
∑F = Ff (Force due to friction)
The net force is the centripetal force, so:
mv²/r = Ff
Rewrite the force due to friction:
mv²/r = μmg
Cancel out the mass:
v²/r = μg
Solve for v:
v = √rμg
v = √(25)(9.81)(0.8) = 14.01 m/s
calculate 18% of 2758 correct to 4 significant figure
Answer:
......the answer is 496.4
In an oscillating LC circuit, when 81.0% of the total energy is stored in the inductor's magnetic field, (a) what multiple of the maximum charge is on the capacitor and (b) what multiple of the maximum current is in the inductor
Answer:
21
Explanation:
9+10=21
A vector has a magnitude of 30m at an angle of 225 degrees with respect to the positive x-axis and has a magnitude of 13m.what are the Component of vectors B and it's direction
Answer:
I'm pretty sure this is not a complete question. My guess is that you are trying to add/subtract vectors. Vectors have both magnitude and direction, so vector A is pretty clear, but a magnitude of 13 (i'm guessing a resultant) without a direction is weird.
IF 13 is the magnitude of the resultant, vector B added to vector A could have any magnitude 17 ≤ B ≤ 43
It could have any direction of
θ = (225 - 180) ± arcsin(13/30)
θ = 45 ± 25.679...
70.679 ≤ θ ≤ 19.321
components of vector B would be
Bx = |B|cosθ
By = |B|sinθ
My guess is that you are trying to add/subtract vectors. Vectors have both magnitude and direction, so vector A is pretty clear, but a magnitude of 13 (i'm guessing a resultant) without a direction is weird.
What is magnitude of the resultant?IF 13 is the magnitude of the resultant, vector B added to vector A could have any magnitude 17 ≤ B ≤ 43
It could have any direction of
θ = (225 - 180) ± arcsin(13/30)
θ = 45 ± 25.679...
70.679 ≤ θ ≤ 19.321
components of vector B would be
Bx = |B|cosθ
By = |B|sinθ
My guess is that you are trying to add/subtract vectors. Vectors have both magnitude and direction, so vector A is pretty clear, but a magnitude of 13 (i'm guessing a resultant) without a direction is weird. A vector has a magnitude of 30m at an angle of 225 degrees with respect to the positive x-axis and has a magnitude of 13m.
Therefore, My guess is that you are trying to add/subtract vectors. Vectors have both magnitude and direction, so vector A is pretty clear, but a magnitude of 13 (i'm guessing a resultant) without a direction is weird.
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in vacuum , the shorter the wavelength of an electromagnetic wave is , the:
A. lower its frequency
B. higher its energy
C. longer its period
D. slower its speed
Answer:
Higher its Energy
Explanation:
A 220 g mass is on a frictionless horizontal surface at the end of a spring that has force constant of 7.0
Nm-1
. The mass is displaced 5.2 cm from its equilibrium position and then released to undergo simple
harmonic motion.
At what displacement from the equilibrium position is the potential energy equal to the kinetic energy.
The conservation of energy allows to find the result for the point where the kinetic and potential energy are equal is;
x = 0.026 m
Given parameters
The mass of the body m = 220 g = 0.220 kg The force constant k = 7.0 N / m The initial displacement or amplitude xo = 5.2 cm = 0.052 mTo find
The point where scientific and potential energy are equal.
The law of the conservation of mechanical energy is one of the most important in physics, stable that if there is no friction, the mechanical energy of the system is conserved. The mechanical energy is formed by the sum of the kinetic energy and the potential energies.
Em = K + U
Let's write the energy in two points.
Starting point. With maximum compression.
Em₀ = U = ½ k x²
Final point. Where the kinetic and potential energy are equal.
[tex]Em_f = K +U[/tex]
Since the mechanical energy is constant at this point K = U, therefore we can write the energy.
[tex]Em_f = 2U = 2 ( \frac{1}{2} \ k \ x_f^2 )[/tex]
Energy is conserved.
[tex]Em_o = Em_f \\\frac{1}{2} \ k x_o^2 = 2 ( \frac{1}{2} \ k x_f^2)[/tex]Emo = Emf
½ k x² = 2 (½ k xf²)
[tex]x_f = \frac{x_o}{2}[/tex]
let's calculate.
[tex]x_f = \frac{0.052}{2} \\x_f = 0.026 m[/tex]
In conclusion using the conservation of energy we can find the point where the kinetic and potential energy are equal is;
x = 0.026 m
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