First, determine the mass of the object by dividing its weight on Earth by 9.8 m/s² as shown below,
m = 250 N / 9.8 m/s² = 25.51 kg
Then, multiply the obtained mass by the acceleration due to gravity (g) on Pluto.
W (in Pluto) = (25.51 kg) x (0.61 m/s²) = 15.56 N
Therefore, the object will only weigh 15.56 N.
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That's what the word means.
Answer:
ΔL = 3.82 10⁻⁴ m
Explanation:
This is a thermal expansion exercise
ΔL = α L₀ ΔT
ΔT = T_f - T₀
where ΔL is the change in length and ΔT is the change in temperature
Let's reduce the length to SI units
L₀ = 90.5 mm (1m / 1000 mm) = 0.0905 m
let's calculate
ΔL = 25.10⁻⁶ 0.0905 (154.6 - (14.4))
ΔL = 3.8236 10⁻⁴ m
using the criterion of three significant figures
ΔL = 3.82 10⁻⁴ m
Answer:
U₂ = 400 KJ
Explanation:
Given that
Initial energy of the tank ,U₁= 800 KJ
Heat loses by fluid ,Q= - 500 KJ
Work done on the fluid ,W= - 100 KJ
Sign -
1.Heat rejected by system - negative
2.Heat gain by system - Positive
3.Work done by system = Positive
4.Work done on the system-Negative
Lets take final internal energy =U₂
We know that
Q= U₂ - U₁ + W
-500 = U₂ - 800 - 100
U₂ = -500 +900 KJ
U₂ = 400 KJ
Therefore the final internal energy = 400 KJ
