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Ghella [55]
2 years ago
12

A small lab cart and one of larger mass collide and rebound off each other. Which of them has the greater average force on it du

ring the collision?
They both experience a force of zero.


They both experience the same magnitude of the collision force.


The large cart has greater force on it.


The small cart has greater force on it.
Physics
1 answer:
I am Lyosha [343]2 years ago
7 0

When a small lab cart collide with a large mass then during the collision two bodies will remain in contact and then apply the contact force on each other

This contact force is always equal on two bodies in magnitude and also it is normal or perpendicular to the contact surface.

This will always follow Newton's III law as per which every action has equal and opposite reaction.

So this contact force will not depend on the mass or any other factor but it will remain same on two colliding bodies.

So here the correct answer would be

<em>They both experience the same magnitude of the collision force. </em>


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In the swing carousel amusement park ride, riders sit in chairs that are attached by a chain to a large rotating drum as shown i
irinina [24]

Answer:\theta =44.068^{\circ}

Explanation:

Given

time taken to complete the circle=7.9 s

radius of circle(r)=15 m

velocity of rider is given by =\frac{2\pi r}{t}

v=\frac{2\pi 15}{7.9}=11.93 m/s

Let us suppose T is the tension in the chain and \thetais the angle which chain makes with vertical

Therefore T\sin \theta =\frac{mv^2}{r}-1

T\cos \theta=mg --2

Divide 1 & 2 we get

tan\theta =\frac{v^2}{rg}

tan\theta =0.968

\theta =44.068^{\circ}

8 0
2 years ago
A 96-mH solenoid inductor is wound on a form 0.80 m in length and 0.10 m in diameter. A coil is tightly wound around the solenoi
Sholpan [36]

Answer:

i = 7.83 \mu A

Explanation:

Induced EMF in the coil is given by the equation

EMF = M\frac{di}{dt}

so we have

M = 31 \mu H

also we know that rate of change in current in solenoid is given as

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so induced EMF of coil is given as

EMF = (31 \times 10^{-6})(2.5)

EMF = 77.5 \times 10^{-6} A/s

now induced current in the coil will be given as

i = \frac{EMF}{R}

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i = 7.83 \mu A

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Answer:

yuhhh

Explanation:

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To solve this problem it is necessary to apply the concepts related to Young's Module and its respective mathematical and modular definitions. In other words, Young's Module can be expressed as

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\Delta L= Compression

L_0= Original Length

According to the values given we have to

\Upsilon_{steel} = 200*10^9Pa

\Delta L = 5.6*10^{-7}m

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Replacing this values at our previous equation we have,

\Upsilon = \frac{F/A}{\Delta L/L_0}

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