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If the gravitational pull is zero and I multiply by mass I get a zero
Answer:
onservation of energy
U top = K bottom
(m + m)*g*L = 1/2*I*?^2 where I = m*(L/2)^2 + m*L^2 = 1.25*m*L^2
So 2m*g*L = 1/2*1.25*m*L^2*?^2
So ? = sqrt(3*g*/(1.25*L) ) = sqrt(12g/5L)
Answer:
ΔU = - 310.6 J (negative sign indicates decrease in internal energy)
W = 810.6 J
Explanation:
a.
Using first law of thermodynamics:
Q = ΔU + W
where,
Q = Heat Absorbed = 500 J
ΔU = Change in Internal Energy of Gas = ?
W = Work Done = PΔV =
P = Pressure = 2 atm = 202650 Pa
ΔV = Change in Volume = 10 L - 6 L = 4 L = 0.004 m³
Therefore,
Q = ΔU + PΔV
500 J = ΔU + (202650 Pa)(0.004 m³)
ΔU = 500 J - 810.6 J
<u>ΔU = - 310.6 J (negative sign indicates decrease in internal energy)</u>
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b.
The work done can be simply calculated as:
W = PΔV
W = (202650 Pa)(0.004 m³)
<u>W = 810.6 J</u>
1,3 and 5 are the answers
