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IceJOKER [234]
4 years ago
7

Dimension of young modulus​

Physics
2 answers:
Lisa [10]4 years ago
6 0

Answer:

M/LT^2

Explanation:

Young's modulus is basically the ability of an object to resist change in its length when undergoes a tension or compression. It is a mechanical property of an object.

Now the formula for young's modulus or the modulus of elasticity is:

Young's modulus = Stress/Strain

As we know, strain is a dimensionless quantity, because it is a change in dimension divided by the original dimension, whereas stress has the same dimensions as pressure. Stress is

Stress = Force/Area

Now

Force = ma,

Where m = mass

           a = acceleration

So, Stress = (ma)/Area

Acceleration = Velocity/Time

and Velocity = Distance/Time

So Acceleration = (Distance/Time)/Time

Now, what we have to understand here, is the basic dimensions are

M = mass

L = Length (or distance)

T = Time

So, by putting values for dimensions of stress

Stress (Dimensions) =  Force/Area

Since Area = (meter)^2

Stress (Dimensions) = Force/L^2

substituting values for force

Stress (Dimensions) = ma/L^2

For mass dimension is M

Stress (Dimensions) = Ma/L^2

for acceleration, dimensions are L/T^2

Stress (Dimensions) = M(L/T^2)/L^2

by simplifying:

Stress (Dimensions) = M/LT^2

As we know that dimensions for Young's modulus will be same as the dimensions for Stress, So

Young's Modulus (Dimensions) = M/LT^2

Lostsunrise [7]4 years ago
4 0

Answer:

IDK leave me alone !

Explanation:

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4 0
3 years ago
A bullet of mass 0.0021 kg initially moving at 497 m/s embeds itself in a large fixed piece of wood and travels 0.65 m before co
alina1380 [7]

Answer:

The force exerted by the wood on the bullet is 399.01 N

Explanation:

Given;

mass of bullet, m = 0.0021 kg

initial velocity of the bullet, u = 497 m/s

final velocity of the bullet, v = 0

distance traveled by the bullet, S = 0.65 m

Determine the acceleration of the bullet which is the deceleration.

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