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borishaifa [10]
3 years ago
6

On the moon, the gravitational acceleration is approximately one-sixth that on the surface of the earth. A 5-kg mass is "weighed

" with a beam balance on the surface of the moon.
a) What is the expected reading?
b) If this mass is weighed with a spring scale that reads correctly for standard gravity on earth, what is the reading?
Physics
1 answer:
Rudiy273 years ago
7 0

Answer:

A. 8.175 N

B. 49.05 N

Explanation:

A.

Acceleration due to gravity (moon) = 1/6 * (acceleration due to gravity (earth)

Acceleration due to gravity (earth) = 9.81 m/s2

Acceleration due to gravity (moon) = 9.81/6

= 1.635 m/s

Weight. Fm = acceleration due to gravity * mass

= 1.635 * 5

= 8.175 N

B. Acceleration due to gravity (earth) * mass = Fe

= 9.81 * 5

= 49.05 N

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Answer:

Compression ratio = 24.42

Explanation:

From Thermodynamic,

Adiabatic Equation ⇒ TV^{γ-1} = Constant.

⇒T₁V₁^{γ-1}  = T₂V₂^{γ-1} ..................(1)

Where T₁= initial Temperature, V₁ = Initial Volume, T₂ = final Temperature, V₂ = Final Volume.

Given:  T₁= 18°C we convert to  Kelvin(K) by adding 273.

∴      T₁= 18°C + 273 = 291K

        T₂ = 733° C  also,we convert to  Kelvin(K) by adding 273

 ∴      T₂ = 733° C +273 =1046K

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 ∴ Rearranging equation(1), we have

    (T₁/T₂) = (V₂/V₁)^{γ-1}................(2)

also rearranging equation(2) we have

   (V₁/V₂)^{γ-1} = (T₂/T₁).

Where (V₁/V₂) = Compression ratio.

∴ (V₁/V₂)^{(7/5)-1} =( 1046/291)

simplifying the index in the equation

I.e (7/5)-1 = (7-5 )/5 = 2/5.

(V₁/V₂)^2/5 =(1046/291)^2/5

Multiplying the power on both side of the equation by 5/2.

∴(V₁/V₂)^(2/5)×(5/2) = (1046/291)^(5/2)

⇒ V₁/V₂= (1046/291)^2.5=( 3.59)^2.5

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∴ Compression ratio = 24.42

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Please help me↓
olga2289 [7]

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