A vibrating object is necessary for the production of sound * 1 point True False

The design of a 60.0 cm industrial turntable requires that it has a kinetic energy of 0.250 j when turning at 45.0 rpm. What mus

Aneli [31]

Answer:

The moment of inertia of the turntable about the rotation axis is 0.0225 kg.m²

Explanation:

Given;

radius of the turnable, r = 60 cm = 0.6 m

rotational kinetic energy, E = 0.25 J

angular speed of the turnable, ω = 45 rpm

The rotational kinetic energy is given as;

$E_{rot} = \frac{1}{2} I \omega ^2$

where;

I is the moment of inertia about the axis of rotation

ω is the angular speed in rad/s

$\omega = 45 \frac{rev}{\min} \times \frac{2 \pi \ rad}{1 \ rev} \times \frac{1 \ \min}{60 \ s} \\\\\omega = 4.712 \ rad/s$

$E = \frac{1}{2} I \omega ^2\\\\I = \frac{2E}{\omega ^2} \\\\I = \frac{2 \ \times \ 0.25}{(4.712)^2} \\\\I = 0.0225 \ kg.m^2$

Therefore, the moment of inertia of the turntable about the rotation axis is 0.0225 kg.m²

5 0

3 years ago

Newton's "quantity of motion" is conservation.<br><br> True<br><br> False

Valentin [98]

The answer is false hope this helps

6 0

3 years ago

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A cylinder with initial volume V contains a sample of gas at pressure p. On one end of the cylinder, a piston is let free to mov

Elina [12.6K]

Answer:

The work done on the gas = 2PV

Explanation:

In this question, we need to apply the basic gas laws to determine the word done.

For this question, we need to draw a PV diagram (a pressure-volume diagram), which I have made and attached in the attachment. So please refer to that attachment. I will be using this diagram to solve for the work done on the gas.

So, Please refer to the attachment number 1. where x -axis is of volume and y-axis is of pressure.

As we know that, the work done on the gas is equal to the area under the curve.

W = Area of the triangle

W = 0.5 x ( base) x ( height)

W = 0.5 x (BC) x (AC)

W = 0.5 x (3V-V) x (3P-P)

W = 2PV

Hence, the work done on the gas = 2PV

4 0

3 years ago

In the figure, a weightlifter's barbell consists of two identical small but dense spherical weights, each of mass 50 kg. These w

kondaur [170]

The moment of inertia is $24.8 kg m^2$

Explanation:

The total moment of inertia of the system is the sum of the moment of inertia of the rod + the moment of inertia of the two balls.

The moment of inertia of the rod about its centre is given by

$I_r = \frac{1}{12}ML^2$

where

M = 24 kg is the mass of the rod

L = 0.96 m is the length of the rod

Substituting,

$I_r = \frac{1}{12}(24)(0.96)^2=1.84 kg m^2$

The moment of inertia of one ball is given by

$I_b = mr^2$

where

m = 50 kg is the mass of the ball

$r=\frac{L}{2}=\frac{0.96}{2}=0.48 m$ is the distance of each ball from the axis of rotation

So we have

$I_b = (50)(0.48)^2=11.5 kg m^2$

Therefore, the total moment of inertia of the system is

$I=I_r + 2I_b = 1.84+ 2(11.5)=24.8 kg m^2$

Learn more about inertia:

brainly.com/question/2286502

brainly.com/question/691705

#LearnwithBrainly

6 0

3 years ago

A radiographer stands six feet from the x-ray source when performing a portable chest exam and receives an exposure of 2 mGy. If

telo118 [61]

Answer:

I₂ = 8 mG

Explanation:

The intensity of a beam is

I = P / A

Where P is the emitted power which is 3) 3

Let's use index 1 for the initial position of r₁ = 6 ft and 2 for the second position r₂ = 3 ft

I₁ A₁ = I₂ A₂

I₂ = I₁ A₁ / A₂

The area of the beam if we assume that it is distributed either in the form of a sphere is

A₁ = 4π r²

We substitute

I₂ = I₁ (r₁ / r₂)²

I₂ = 2 (6/3)²

I₂ = 2 4

I₂ = 8 mG

3 0

3 years ago

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