Answer:
90 ft/s is what i put. Let me know if its wrong
Answer:
option C
Explanation:
given,
energy dissipated by the system to the surrounding = 12 J
Work done on the system = 28 J
change in internal energy of the system
Δ U = Q - W
system losses energy = - 12 J
work done = -28 J
Δ U = Q - W
Δ U = -12 -(-28)
Δ U = 16 J
hence, the correct answer is option C
initial speed of the racer is given as
after applied force the final speed is given as
now during this speed change the racer will cover total distance 185 m
so here we will use kinematics
now the force that chute will exert on the racer will be given as
B) here following is the strategy for solving it
1. first we used kinematics to find the acceleration of the car
2. then we used Newton's II law (F = ma) to find the force
Answer:
Power output is greater
Explanation:
The books mentioned are all identical and the to which these books are being lift up is same in both the slow and the fast process.
Now , we know that the work done is the scalar product of force applied and its displacement and displacement in both the cases is same. Thus the net work done is also same.
If we talk about Power, then power is work done per unit time. It includes the time which was absent in work done and also the time here is in inverse proportion to power. Thus slower process will take more time and less power while faster one will take less time and more power.
