Answer:
10 kJ
Explanation:
W = Fd
W = (μN)(vt)
W = μ(mg)vt
W = 0.7(42.9)(9.81)(9)(3.8)
W = 10,075.12506 J
W ≈ 10 kJ
Answer:
a)W= - 720 J
b)ΔU= 330 J
Explanation:
Given that
P = 0.8 atm
We know that 1 atm = 100 KPa
P = 80 KPa
V₁ = 12 L = 0.012 m³ ( 1000 L = 1 m³)
V₂ = 3 L = 0.003 m³
Q= - 390 J ( heat is leaving from the system )
We know that work done by gas given as
W = P (V₂ -V₁ )
W= 80 x ( 0.003 - 0.012 ) KJ
W= - 0.72 KJ
W= - 720 J ( Negative sign indicates work done on the gas)
From first law of thermodynamics
Q = W + ΔU
ΔU=Change in the internal energy
Now by putting the values
- 390 = - 720 + ΔU
ΔU= 720 - 390 J
ΔU= 330 J
Answer:
520000 or 520000 pa
Force = 520N
Area of contact = 0.001
Pressure: 520000 or 520000
The question doesn't describe any experiment. If the same experiment is repeated, no matter how many times, the acceleration due to gravity will remain the same as it was during the non-existent original experiment, and will have no effect on anything.
