It takes sunlight 8 minutes to reach earth , so yes
The mechanical work done by the sprinter during this time will be 4537.5 J , the average power the sprinter must generate will be 907.5 W and if the sprinter converts food energy to mechanical energy with an efficiency of 25% then he will be burning calories at 54.20 calories per second.
Work in physics is the energy that is transferred to or from an item when a force is applied along a displacement. It is frequently described in its most basic form as the result of force and displacement.
The quantity of energy moved or transformed per unit of time is known as power in physics. The watt, or one joule per second, is the unit of power in the International System of Units.. A scalar quantity is power.
Given 75-kg sprinter accelerates from rest to a speed of 11.0 m/s in 5.0 s.
So let,
m = 75 kg
v = 11.0 m/s
t = 5.0 s
So the mechanical work done by the sprinter during this time will be as follow:
W = 0.5 mv²
W = 0.5 (75)(11)²
W = 4537.5 J
The average power the sprinter must generate will be as follow:
Power(P) = W / t
P = 4537.5/5
P = 907.5 W
Only 25% is absorbed. So, the sprinter only absorbed 226.875 J per second which is equal to 54.20 calories per second.
Hence mechanical work done by the sprinter during this time will be 4537.5 J , the average power the sprinter must generate will be 907.5 W and if the sprinter converts food energy to mechanical energy with an efficiency of 25% then he will be burning calories at 54.20 calories per second.
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The acceleration of the car is solved by subtracting the initial speed from the final speed then dividing the result by the elapsed time.
initial speed = 72 km/hr = 20 m/s
final speed = 0 m/s
elapsed time = 5 seconds
acceleration = (0 m/s – 20 m/s) / 5 s
acceleration = - 20m/s / 5 s
acceleration = -4 m/s^2
need speed of sound on lhs
There are missing data in the text of the problem (found them on internet):
- speed of the car at the top of the hill:

- radius of the hill:

Solution:
(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car

(downwards) and the normal force N exerted by the road (upwards). The resultant of these two forces is equal to the centripetal force,

, so we can write:

(1)
By rearranging the equation and substituting the numbers, we find N:

(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:

(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:

from which we find