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
A theromometer is the increase or decrease of earths atmospheric temperture, thats how you would measure the temperture of the air around you.
Answer:
a) 6.4 kJ
b) 43.4 kJ
Explanation:
a)
= Heat absorbed = 37 kJ
= Coefficient of performance = 5.8
= Work done
Heat absorbed is given as
=
37 = (5.8)
= 6.4 kJ
b)
= work per cycle required
= +
= 37 + 6.4
= 43.4 kJ
Answer:
a) -1.25 rev/s² and 23.3 rev
b) 2.67s
Explanation:
a) ω
= (500 rev/min)(1min/ 60s) => 8.333 rev/s
ω
= (200 rev/min)(1min/ 60s) => 3.333rev/s
time 't'= 4 s
angular acceleration 'α'=?
constant angular acceleration equation is given by,
ω= ω + αt
α= (ω - ω )/t => (3.333-8.333)/4
α= -1.25 rev/s²
θ-θ
= ω t + 1/2αt²
=(8.333)(4) + 1/2 (-1.25)(4)²
=23.3 rev
b) ω=0 (comes to rest)
ω = 3.333 rev/s
α= -1.25 rev/s²
ω= ω + αt
t= (ω - ω)/α => (0- 3.333)/-1.25
t= 2.67s
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I'm not really sure but I do know that it's not 0 because the object is still moving, even if it's only moving at 1 m/s.
