Answer:
a) -41.1 Joule
b) 108.38 Kelvin
Explanation:
Pressure = P = 290 Pa
Initial volume of gas = V₁ = 0.62 m³
Final volume of gas = V₂ = 0.21 m³
Initial temperature of gas = T₁ = 320 K
Heat loss = Q = -160 J
Work done = PΔV
⇒Work done = 290×(0.21-0.62)
⇒Work done = -118.9 J
a) Change in internal energy = Heat - Work
ΔU = -160 -(-118.9)
⇒ΔU = -41.1 J
∴ Change in internal energy is -41.1 J
b) V₁/V₂ = T₁/T₂
⇒T₂ = T₁V₂/V₁
⇒T₂ = 320×0.21/0.62
⇒T₂ = 108.38 K
∴ Final temperature of the gas is 108.38 Kelvin
Harmonics, Loop and Harmonic number
Hope this helps :)
Answer:
Explanation:
Given:
U1 = 1.6 m/s
U2 = -1.1 m/s
M1 = 1850 kg
M2 = 1400 kg
V1 = 0.27 m/s
Using momentum- collision equation,
M1U1 + M2U2 = M1V1 + M2V2
1850 × 1.6 - 1400 × 1.1 = 1850 × 0.27 + 1400 × V2
1420 = 499.5 + 1400V2
V2 = 0.6575 m/s
B.
KE = 1/2 × MV^2
KEa1 + KEa2 = KEb1 + KEb2
Delta KE = KE2 - KE1
KEa1 = 2368 J
KEb1 = 847 J
KEa2 = 67.433 J
KEb2 = 302.6 J
KE1 = KEa1 + KEb1
= 3215 J
KE2 = 370.033 J
Delta KE = -2845 J.
