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Rufina [12.5K]
3 years ago
15

Determine the increase in volume of 100m3 of mercury when it's temperature change from 10c to 45c the linear expansion coefficie

nt is 6 x 10-5
Physics
1 answer:
alisha [4.7K]3 years ago
7 0

Answer:

0.63m

Explanation:

Volume expansivity = change in volume/original volume×temp change

Volume expansivity p = 3x

p = ∆V/V∆t

x is the linear expansivity

Given

x = 6 x 10^-4

p = 3x

p = 3(6 x 10^-5)

p = 18×10^-5

Volume = 100m³

∆t = 45-10 = 35°C

Required

Change in volume ∆V

Substitute the given values into the formula

18×10^-5 = ∆V/100(35)

18×10^-5 =∆V/3500

∆V = 3500×0.00018

∆V = 0.63m

Hence the increase in volume of the Mercury is 0.63n

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Answer:

0.0327 m

Explanation:

m = 2 kg

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Equate both the energies as according to the question

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An asteroid orbits the Sun every 176 years. What is the asteroids average distance from the Sun? P ^ 2 = a ^ 3 where p = period
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Answer:

The value is  x =  45.99 \  Au

Explanation:

From the question we are told that

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Generally the average distance of the asteroid from the sun is mathematically represented as

            R = \sqrt[3]{ \frac{G M * T^2 }{4 \pi} }

Here M is the mass of the sun with a value  

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           R = \sqrt[3]{ \frac{6.67 *10^{-11}  * 1.99*10^{30} * [5.55 *10^{9}]^2 }{4 * 3.142 } }

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          R = 6.88 *10^{12} \  m \ \ \ \ \to \ \   x \  Au

=>      x =  \frac{6.88 *10^{12}}{1.496 *10^{11}}

=>       x =  45.99 \  Au

       

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