A meter stick is found to balance at the 49.7-cm mark when placed on a fulcrum. When a 65.5-gram mass is attached at the 21.0-cm
1 answer:
Answer:
113.53 g
Explanation:
Please see attached photo for explanation.
In the attached photo, M is the mass of the meter stick.
The value of M can be obtained as shown below:
Clockwise moment = M × 10.5
Anticlockwise moment = 65.5 × 18.2
Anticlockwise moment = Clockwise moment
65.5 × 18.2 = M × 10.5
1192.1 = M × 10.5
Divide both side by 10.5
M = 1192.1 / 10.5
M = 113.53 g
Thus, the mass of the meter stick is 113.53 g
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Answer:
57,42 KJ
Explanation:
By a isobaric proces, the expresion for the works in the jpg adjunt. Then:
W = Pa(Vb - Va) = Pa*Vb - Pa*Va ---(1)
By the ideal gases law: PV=RTn
Then, in (1): (remember Pa = Pb)
W = R*Tb*n - R*T*an = R*n*(Tb - Ta) --- (2)
Since we have 1 Kg air: How much is this in moles?
From bibliography: 28.96 g/mol
Then, in 1 Kg (1000 g) there are:
n = 34,53 mol
Finally, in (2):
W = (8,3144 J/K.mol)*(34,53 mol)*(500K - 300K) = 51 419,9 J ≈ 57,42 KJ
