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Allisa [31]
3 years ago
11

A copper rod of length 27.5 m has its temperature increases by 35.9 degrees celsius. how much does its length increase?(unit=m)

Physics
2 answers:
sineoko [7]3 years ago
6 0

Answer:0.01678325m

Explanation:

Original length(L1)=27.5m

Temperature rise(@)=35.9°C

Linear expansivity(α)=0.000017

Length increase(L)=?

L=α x L1 x @

L=0.000017 x 27.5 x 35.9

L=0.01678325m

gavmur [86]3 years ago
5 0
<h2>The increase in length = 1.87 x 10⁻²</h2>

Explanation:

When copper rod is heated , its length increases

The increase in length can be found by the relation

L = L₀ ( 1 + α ΔT )

here L is the increased length and L₀ is the original length

α  is the coefficient of linear expansion and ΔT is the increase in temperature .

The increase in length = L - L₀ = L₀ x α ΔT

Substituting all these value

Increase in length = 27.5 x 1.7 x 10⁻⁵ x 35.9

= 1.87 x 10⁻² m

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Answer:

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Explanation:

The question parameters are;

The maximum long jump distance of a person on Earth, R_{max} = 3 m

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The distance the person can jump on the Moon is given as follows;

A person performing a jump across an horizontal distance on Earth (under gravitational force) follows the path of the motion of a projectile

The horizontal range, R_{max}, of a projectile motion is found by using the following formula

R_{max} = \dfrac{u^2}{g}

Where;

g = The acceleration due to gravity = 9.8 m/s²

Therefore, we have;

R_{max} = 3 \, m = \dfrac{u^2}{9.8 \, m/s^2 }

u² = 3 m × 9.8 m/s² = 29.4 m²/s²

Therefore, on the Moon, we have;

The acceleration due to gravity on the Moon, g_{Moon} = 1/16 × g

∴ g_{Moon} = 1/16 × g = 1/16 × 9.8 m/s² ≈ 0.6125 m/s²

R_{max \ Moon} = \dfrac{u^2}{g_{Moon}}   = \dfrac{29.4 \ m^2/s^2}{0.6125 \, m/s^2 } \approx 48 \, m

The maximum distance the person can jump on the Moon with the same velocity which was used on Earth is R_{max \ Moon} ≈ 48 m

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