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Mars2501 [29]
3 years ago
5

A energy storage system based on a flywheel (a rotating disk) can store a maximum of 4.0 MJ when the flywheel is rotating at 20,

000 revolutions per minute. What is the moment of inertia of the flywheel?
Physics
1 answer:
mariarad [96]3 years ago
8 0

Answer:

Moment of inertia of the flywheel, I=1.82\ kg-m^2          

Explanation:

Given that,

The maximum energy stored on  flywheel, E=4\ MJ=4\times 10^6\ J

Angular velocity of the flywheel, \omega=20000\ rev/s=2094.39\ rad/s

We need to find the moment of inertia of the flywheel. The energy of a flywheel in rotational kinematics is given by :

E=\dfrac{1}{2}I\omega^2

I is the moment of inertia of the flywheel

On rearranging we get :

I=\dfrac{2E}{\omega^2}

I=\dfrac{2\times 4\times 10^6}{(2094.39)^2}

I=1.82\ kg-m^2

So, the moment of inertia of the flywheel is I=1.82\ kg-m^2. Hence, this is the required solution.

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Answer:

Bone

Explanation:

Diagnostic radiology include the use of non-invasive imaging scans to diagnose a patient.

The voltages used in diagnostic tubes range from roughly 20 kV to 150 kV and thus the highest energies of the X-ray photons range from roughly 20 keV to 150 keV.

The tests and equipment used sometimes involves low doses of radiation to create highly detailed images of an area.

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Answer:

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3 years ago
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A 2.45-kg frictionless block is attached to an ideal spring with force constant 355 N/m. Initially the spring is neither stretch
ANTONII [103]

Answer:

A.    A = 0.913 m

B.    amax = 132.24m/s^2

C.    Fmax = 324.01N

Explanation:

When the block is moving at the equilibrium point , its velocity is maximum.

A. To find the amplitude of the motion you use the following formula for the maximum velocity:

v_{max}=A\omega          (1)

vmax = maximum velocity = 11.0 m/s

A: amplitude of the motion = ?

w: angular frequency = ?

Then, you have to calculate the angular frequency of the motion, by using the following formula:

\omega=\sqrt{\frac{k}{m}}           (2)

k: spring constant = 355 N/m

m: mass of the object = 2.54 kg

\omega = \sqrt{\frac{355N/m}{2.45kg}}=12.03\frac{rad}{s}

Next, you solve the equation (1) for A and replace the values of vmax and w:

A=\frac{v}{\omega}=\frac{11.0m/s}{12.03rad/s}=0.913m

The amplitude of the motion is 0.913m

B. The maximum acceleration of the block is given by:

a_{max}=A\omega^2 = (0.913m)(12.03rad/s)^2=132.24\frac{m}{s^2}

The maximum acceleration is 132.24 m/s^2

C. The maximum force is calculated by using the second Newton law and the maximum acceleration:

F_{max}=ma_{max}=(2.45kg)(132.24m/s^2)=324.01N

It is also possible to calculate the maximum force by using:

Fmax = k*A = (355N/m)(0.913m) = 324.01N

The maximum force exertedbu the spring on the object is 324.01 N

4 0
3 years ago
A 0.171-kilogram hockey puck is struck by a hockey stick. If the force from the hockey stick is 30.2 Newtons, but the puck also
Komok [63]

Answer:

a = 156.14 m/s^2

Explanation:

Using the laws of newton:

∑F = ma  

where ∑F is the sumatory of forces, m the mass and a the aceleration.

so:

F -f_k = ma

where F is the force from the hockey stick and f_k is the force of friction.

Replacing values, we get:

(30.2N) - (3.5N)= (0.171kg)a

Finally, solving for a:

a = 156.14 m/s^2

4 0
3 years ago
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