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GREYUIT [131]
3 years ago
5

What is the moment of inertia of a disc of mass 5kg and radius 10cm?

Physics
2 answers:
vodka [1.7K]3 years ago
6 0

Answer:

500

Explanation:

Lelu [443]3 years ago
3 0
I=1/2MR²
I=1/2(5 kg)(0.1 m)²
I=0.025 kg/m²
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A positive test charge q is released from rest at distance r away from a charge of Q and a distance 2r away from a charge of 2Q.
Luba_88 [7]

Answer: Option (b) is the correct answer.

Explanation:

It is given that a positive test charge q is released from rest at a distance r away from a charge of +Q and a distance 2r which is away from a charge of +2Q.

Then test charge to the right immediately after being released.

Therefore, the net force will be as follows.

            F = \frac{kqQ}{r^{2}} - kq\frac{(2Q)}{(2r)^{2}}

               = \frac{4KqQ - 2KqQ}{4r^{2}}

               = \frac{KqQ}{2r^{2}}

           F = \frac{KqQ}{2r^{2}} > 0

Thus, we can conclude that the test charge move to the right immediately after being released.

7 0
3 years ago
A 60 cm diameter wheel accelerates from rest at a rate of 7 rad/s2. After the wheel has undergone 14 rotations, what is the radi
Mamont248 [21]

Answer:

a=368.97\ m/s^2

Explanation:

Given that,

Initial angular velocity, \omega=0

Acceleration of the wheel, \alpha =7\ rad/s^2

Rotation, \theta=14\ rotation=14\times 2\pi =87.96\ rad

Let t is the time. Using second equation of kinematics can be calculated using time.

\theta=\omega_it+\dfrac{1}{2}\alpha t^2\\\\t=\sqrt{\dfrac{2\theta}{\alpha }} \\\\t=\sqrt{\dfrac{2\times 87.96}{7}} \\\\t=5.01\ s

Let \omega_f is the final angular velocity and a is the radial component of acceleration.

\omega_f=\omega_i+\alpha t\\\\\omega_f=0+7\times 5.01\\\\\omega_f=35.07\ rad/s

Radial component of acceleration,

a=\omega_f^2r\\\\a=(35.07)^2\times 0.3\\\\a=368.97\ m/s^2

So, the required acceleration on the edge of the wheel is 368.97\ m/s^2.

6 0
3 years ago
Panel A shows a ball shortly after being thrown upward. Panel B shows the same ball in an instant on its way down. Suppose air r
Anna11 [10]

Answer:

ur mom

Explanation:

6 0
2 years ago
Football player 1 has a mass of 80 kg and a velocity of 2 m/s east while player 2 has a mass of 70 kg and a velocity of 3 m/s we
sukhopar [10]

The total momentum of the system is equal to 50 Kgm/s.

<u>Given the following data:</u>

  • Mass 1 = 80 kg
  • Velocity 1 = 2 m/s east.
  • Mass 2 = 70 kg
  • Velocity 2 = 3 m/s west.

To determine the total momentum of the system:

Mathematically, momentum is given by the formula;

Momentum = mass \times velocity

<u>For Football player 1:</u>

Momentum = 80 \times 2

Momentum 1 = 160 Kgm/s.

<u>For Football player 2:</u>

Momentum = 70 \times 3

Momentum 1 = 210 Kgm/s.

Now, we can calculate the total momentum of the system:

Total\;momentum = momentum \;2 - momentum \;1\\\\Total\;momentum = 210-160

Total momentum = 50 Kgm/s.

<u>Note:</u> We subtracted because the football players were moving in opposite directions.

Read more: brainly.com/question/15517471

6 0
2 years ago
A rope.that is 1.50m long exhibits a standing wave pattern with 6 nodes if the frequency of the waves is 75.0 hz what is the vel
Sonja [21]
I think he answer is 34.32
4 0
3 years ago
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