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3 years ago

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You hang a heavy ball with a mass of 42 kg from a silver rod 2.7 m long by 1.9 mm by 2.6 mm. You measure the stretch of the rod,

and find that the rod stretched 0.002902 m. Using these experimental data, what value of Young's modulus do you get

1 answer:

nadezda [96]3 years ago

5 0

Answer:

Explanation:

cross sectional area A = 1.9 x 2.6 x 10⁻⁶ m²

= 4.94 x 10⁻⁶ m²

stress = 42 x 9.8 / 4.94 x 10⁻⁶

= 83.32 x 10⁶ N/m²

strain = .002902 / 2.7

= 1.075 x 10⁻³

Young's modulus = stress / strain

= 83.32 x 10⁶ / 1.075 x 10⁻³

= 77.5 x 10⁹ N/m²

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Explanation:

From the question we are told that

The temperature is $T = 7200K$

The diameter of the ball is $d = 1.5 km = 1.5 *1000 = 1500m$

Hence the radius = $= \frac{1500}{2} = 750m$

The total energy radiated can be mathematically represented as

$E = \sigma A T^4$

Where $\sigma$ is the Stefan-Boltzmann constant $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 5.67*10^{-8} W \ \cdot m^{-2} K^{-4}$

A is the area of a sphere = $\pi d^2 = 3.142 * 1500^2 = 7.069500 *10 ^6\ m^2$

Substituting values we have

$E = 5,67*10^{-8} * 7.069500*10^6 * 7200^4$

$=1.077*10^{15} W$

Now the state of the energy is mathematically represented as

$Rate \ of \ energy \ radiation (E_r)= \frac{E}{A} = \sigma T^4$

$= 5.67*10^{-8} * 7200^2$

$= 1.523747635*10^9 W/m^2$

A sketch illustrating the b part of the question is shown on the first uploaded image

looking at the height at which the blast occurs(16km) as compared to the height of the wall we notice that the height of the wall is negligibly small

from the diagram x can be calculated as follows

$x = \sqrt{40^2 + 16^2}$

$= 43.0813 Km$

This value of x represents the radius of the blast(assuming it is spherical ) when it is at that wall

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$G = \frac{E}{4 \pi r^2}$

Here r = 43.0813 Km = 43.0813 × 1000 = 43081.3 m

$G= \frac{1.077*10^15}{4 \pi (431081.3^2)}$

$G =46.177\ kW/m^2$

Generally the amount of energy absorbed can be mathematically represented as

$Amount \ of \ energy \ absorbed \ (E_B ) = G * t$

Where t is the time taken

Therefore $E_B = 46.177 *10 = 461.77 KJ$

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