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kvv77 [185]
3 years ago
10

A cylindrical, 0.500-m rod has a diameter of 0.02 m. The rod is stretched to a length of 0.501 m by a force of 3000 N. What is t

he Young's modulus of the material?
Physics
1 answer:
Salsk061 [2.6K]3 years ago
6 0

Answer:

Y = 4.775 x 10⁹ Pa = 4.775 GPa

Explanation:

First, we calculate the stress on the rod:

stress = \frac{Force}{Area} = \frac{3000\ N}{\pi r^2}  \\\\stress = \frac{3000\ N}{\pi (0.01\ m)^2}\\\\stress = 9.55\ x\ 10^6\ Pa = 9.55 MPa\\

Now, we calculate the strain:

strain = \frac{Change\ in Length}{Original\ Length}\\\\strain = \frac{0.501\ m - 0.5\ m}{0.5\ m}\\\\strain =  0.002\\

Now, we will calculate the Young's Modulus (Y):

Y = \frac{stress}{strain}\\\\Y = \frac{9.55\ x\ 10^6\ Pa}{0.002} \\

<u>Y = 4.775 x 10⁹ Pa = 4.775 GPa</u>

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1 year ago
A constant torque is applied to a rigid wheel whose moment of inertia is 2.0 kg · m2 around the axis of rotation. If the wheel s
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Answer:

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Explanation:

Given that,

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The relation between moment of inertia and torque is given by :

\tau=I\alpha \\\\\tau=I\times \dfrac{\omega_f}{t}\\\\\tau=2\times \dfrac{25}{13}\\\\\tau=+3.84\ N-m

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4 0
3 years ago
Physics question, answer completely with work
Alenkasestr [34]

The force is 2.0 N east

Explanation:

The impulse exerted by a force is defined as the product between the force itself and the time interval during which the force is applied. Mathematically, it is equal to the change in momentum experienced by the object on which the force is acting:

I=F\Delta t = \Delta p

Where

I is the impulse

F is the force

Delta t is the time interval during which the force is applied

\Delta p is the change in momentum

In this problem,

\Delta t = 3.0 s is the time interval

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Therefore, the magnitude of the force is

F=\frac{I}{\Delta t}=\frac{6.0}{3.0}=2.0 N

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Arte-miy333 [17]

Answer:

Yes

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The possible values are two: 1045Hz and 1055Hz

3 0
3 years ago
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