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grigory [225]
2 years ago
12

A 12.0-g cd with a radius of 5.81 cm rotates with an angular speed of 33.2 rad/s. what is its kinetic energy?

Physics
1 answer:
schepotkina [342]2 years ago
4 0
Use the formula 1/2*I*w^2 where w is the angular speed and I is the rotaional inirtia (for disk it is .5*mass*radius^2).
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E = 1/2 * m * v²

v = √(2 * E  / m)
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the force between your feet and the floor is greater while standing on your tiptoes than while standing flat on your feet
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Yes thats correct....becuase all of your weight is concentrated on a small area compared to the larger surface area of your feet!
is that what your question was?
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The sensor in the torso of a crash test dummy records the magnitude and direction of the net force acting on the dummy.If the du
Sunny_sXe [5.5K]

Here in crash test the two forces are acting on the dummy in two different directions

As we know that force is a vector quantity so we need to use vector addition laws in order to find the resultant force on it.

So here two forces are given in perpendicular direction with each other so as per vector addition law we need to use Pythagoras theorem to find the resultant of two vectors

so we can say

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here given that

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now we will plug in all data in the above equation

F_{net} = \sqrt{4500^2 + 130^}

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A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3. 0 m is initially at rest. A 20 kg boy approaches the
Margaret [11]

Hi there!

\boxed{\omega = 0.38 rad/sec}

We can use the conservation of angular momentum to solve.

\large\boxed{L_i = L_f}

Recall the equation for angular momentum:

L = I\omega

We can begin by writing out the scenario as a conservation of angular momentum:

I_m\omega_m + I_b\omega_b = \omega_f(I_m + I_b)

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\omega_m = angular velocity of merry go round (rad/sec)

\omega_f = final angular velocity of COMBINED objects (rad/sec)

I_b = moment of inertia of boy (kgm²)

\omega_b= angular velocity of the boy (rad/sec)

The only value not explicitly given is the moment of inertia of the boy.

Since he stands along the edge of the merry go round:

I = MR^2

We are given that he jumps on the merry-go-round at a speed of 5 m/s. Use the following relation:

\omega = \frac{v}{r}

L_b = MR^2(\frac{v}{R}) = MRv

Plug in the given values:

L_b = (20)(3)(5) = 300 kgm^2/s

Now, we must solve for the boy's moment of inertia:

I = MR^2\\I = 20(3^2) = 180 kgm^2

Use the above equation for conservation of momentum:

600(0) + 300 = \omega_f(180 + 600)\\\\300 = 780\omega_f\\\\\omega = \boxed{0.38 rad/sec}

8 0
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Answer: Hi!

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Hope this helps!

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