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3 years ago

Imagine that you’re observing a collision. Which action would allow you to determine whether the collision is inelastic?

Physics

2 answers:

Vesna [10]3 years ago

4 0

Answer:

Observe whether the colliding objects change their shape

Explanation:

I got this question on a test and this was the answer

Thepotemich [5.8K]3 years ago

3 0

The answer is B;) in plato

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How much kinetic energy does an 80 kg man have while running at 3 m/s?

Soloha48 [4]

Hello!

$\large\boxed{KE = 360 J}$

Use the equation KE = 1/2mv² to solve for the kinetic energy of the man.

We are given the mass and velocity, so plug these values into the equation:

KE = 1/2(80)(3²)

KE = 1/2(720)

KE = 360 J

7 0

3 years ago

Everything we know about stars comes from studying their _____.

serious [3.7K]

I think its radiation

3 0

4 years ago

Which of the following is NOT a force that influences the wind? A. coriolis effect B. magnetic field C. pressure gradient D. fri

Pavlova-9 [17]

A magnetic field does not have any influence on wind. The Coriolis effect, pressure gradient, and friction do affect wind.

The solution is B.

7 0

4 years ago

Read 2 more answers

A girl (mass M) standing on the edge of a frictionless merry-go-round (radius R, rotational inertia I) that is not moving. She t

vladimir1956 [14]

a) $\omega=\frac{-mvR}{I+MR^2}$

b) $v=\frac{-mvR^2}{I+MR^2}$

Explanation:

Since there are no external torques acting on the system, the total angular momentum must remain constant.

At the beginning, the merry-go-round and the girl are at rest, so the initial angular momentum is zero:

$L_1=0$

Later, after the girl throws the rock, the angular momentum will be:

$L_2=(I_M+I_g)\omega +L_r$

where:

$I$ is the moment of inertia of the merry-go-round

$I_g=MR^2$ is the moment of inertia of the girl, where

M is the mass of the girl

R is the distance of the girl from the axis of rotation

$\omega$ is the angular speed of the merry-go-round and the girl

$L_r=mvR$ is the angular momentum of the rock, where

m is the mass of the rock

v is its velocity

Since the total angular momentum is conserved,

$L_1=L_2$

So we find:

$0=(I+I_g)\omega +mvR\\\omega=\frac{-mvR}{I+MR^2}$

And the negative sign indicates that the disk rotates in the direction opposite to the motion of the rock.

The linear speed of a body in rotational motion is given by

$v=\omega r$

where

$\omega$ is the angular speed

r is the distance of the body from the axis of rotation

In this problem, for the girl, we have:

$\omega=\frac{-mvR}{I+MR^2}$ is the angular speed

$r=R$ is the distance of the girl from the axis of rotation

Therefore, her linear speed is:

$v=\omega R=\frac{-mvR^2}{I+MR^2}$

5 0

3 years ago

If a construction worker drops a wrench

Svetach [21]

To find velocity at a given moment, also known as instantaneous velocity, we need to know how to take the derivative of an equation;
you can take the derivative of s(t)=62t^2+ 375 by first moving the 2 in front of 162 and multiply both which well give you 324t and 375 will turn to 0
-16t^2 results in -32t
so our velocity at t=4 is -32(4) = -128ft per second

3 0

3 years ago