1) Each of the boxes are pulled 10 m across a level, frictionless floor by a for
ID: 1477730 • Letter: 1
Question
1) Each of the boxes are pulled 10 m across a level, frictionless floor by a force given. Which bod experiences the greatesst change in its kinetic energy?
10 N with 1.0 kg mass
15 N with 2.0 kg mass
5.0 N with 0.50 kg mass
20 N with 3.0 kg mass
12 N with 0.30 kg mass
2) At room temperature, a typical person loses energy to the surroundings at the rate of 62 W. If this energy loss has to be made up by an equivalent food intake, how many kilocalories (food calories) does this person need to consume every day just to make up this heat loss? (1 cal= 4.186 J)
3) What is the best possible performance coefficient (COP) of a heat pump which is used to cool down a fridge to a temperature of 280 K if the room temperature is 292 K?
4) The weight of a 1200 kg car is supported equally by the four tires, which are inflated to the same gauge pressure. What gauge pressure is required so the area of contact of each tire with the road is 100cm^2? (100cm^2=0.01m^2)
5) How many air molecules are in a balloon? (V=0.07 m^2, T=292 K, p=130 kPa)
Explanation / Answer
using work energfy theorem
work done by the net force = change in KE
F*S = dKE
For 10 N with 1.0 kg mass
change in KE = dKE = F*S = 10*10 =100 J
For 15 N with 2.0 kg mass
dKE = 15*10 = 150 J
5.0 N with 0.50 kg mass
dKE = 5*10 = 50 J
20 N with 3.0 kg mass
dKE = 20*10 = 200 J
12 N with 0.30 kg mass
dKE = 12*10 = 120 J
So greatest change in kinetic energy is 20 N with 3.0 kg mass
2) heat loss = power *time = 62*24*60*60 = 5356800 J
but 1 cal = 4.186 J
1J = 1/4.186 cal = 0.238 cal
then required energy consume is 5356800*0.238 = 1279694.21 cal = 1279.69 kcal
3) COP = Th/(Th-Tc) = 292/(292-280) =24.33
4) guage pressure is P = F/A = (1200*9.81/4)/(100*10^-4) = 294300 Pa
5) using P*V = n*R*T
130000*0.07 = n*8.314*292
no.of moles n = 3.74 moles
the no.of air molecules are 3.74*6.022*10^23 = 2.257*10^24 molecules
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