During a demonstration of the gravitational force on falling objects to her class, Sarah drops an 11 lb. bowling ball from the t
1 answer:
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Answer:
(A) 374.4 J
(B) -332.8 J
(C) 0 J
(D) 41.6 J
(E) 351.8 J
Explanation:
weight of carton (w) = 128 N
angle of inclination (θ) = 30 degrees
force (f) = 72 N
distance (s) = 5.2 m
(A) calculate the work done by the rope
- work done = force x distance x cos θ
- since the rope is parallel to the ramp the angle between the rope and
the ramp θ will be 0
work done = 72 x 5.2 x cos 0
work done by the rope = 374.4 J
(B) calculate the work done by gravity
- the work done by gravity = weight of carton x distance x cos θ
- The weight of the carton = force exerted by the mass of the carton = m x g
- the angle between the force exerted by the weight of the carton and the ramp is 120 degrees.
work done by gravity = 128 x 5.2 x cos 120
work done by gravity = -332.8 J
(C) find the work done by the normal force acting on the ramp
- work done by the normal force = force x distance x cos θ
- the angle between the normal force and the ramp is 90 degrees
work done by the normal force = Fn x distance x cos θ
work done by the normal force = Fn x 5.2 x cos 90
work done by the normal force = Fn x 5.2 x 0
work done by the normal force = 0 J
(D) what is the net work done ?
- The net work done is the addition of the work done by the rope, gravitational force and the normal force
net work done = 374.4 - 332.8 + 0 = 41.6 J
(E) what is the work done by the rope when it is inclined at 50 degrees to the horizontal
- work done by the rope= force x distance x cos θ
- the angle of inclination will be 50 - 30 = 20 degrees, this is because the ramp is inclined at 30 degrees to the horizontal and the rope is inclined at 50 degrees to the horizontal and it is the angle of inclination of the rope with respect to the ramp we require to get the work done by the rope in pulling the carton on the ramp
work done = 72 x 5.2 x cos 20
work done = 351.8 J
Answer:
e. TA>T>Tc
Explanation:
a) In this case, we cannot say for sure QA>QB>QC. This is because the magnitude of the heat flow will depend on the specific heat and the mass of each sample. Due to the equation:
if we did an energy balance of the system, we would get that>
QA+QB+QC=0
For this equation to be true, at least one of the heats must be negative. And one of the heats must be positive.
We don't know either of them, so we cannot determine if this statement is true.
b) We can say for sure that QA<0, because when the two samples get to equilibrum, the temperatrue of A must be smaller than its original temperature. Therefore, it must have lost heat. But we cannot say for sure if QB<0 because sample B could have gained or lost heat during the process, this will depend on the equilibrium temperature, which we don't know. So we cannot say for sure this option is correct.
c) In this case we don't know for sure if the equilibrium temperature will be greater or smaller than TB. This will depend on the mass and specific heat of the samples, just line in part a.
d) is not complete
e) We know for sure that A must have lost heat, so its equilibrium temperature must be smaller than it's original temperature. We know that C must have gained heat, therefore it's equilibrium temperature must be greater than it's original temperature, so TA>T>Tc must be true.