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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Answer:
C
Explanation:
Usually when you are at the bottom you are at peak speed. It also shows that Kinetic Energy is the green bar and in picture C the green bar is highest.
Given the following information we have 20 watermelons from mark and 10 fishes from kim therefore we add the longitude of Walmart to the latitude of sams club and end up with a total of 1,000 dish soaps then we convert that into inches which leaves us at 20,000,000 inches of cats then multiply that number to 10 giraffes and we get
1.989 × 10^30 kg and therefore the mass of the sun is 1.989 × 10^30 kg.
Answer:
The bit take to reach its maximum speed of 8,42 x10^4 rad/s in an amount of 1.097 seconds.
Explanation:
ω1= 1.72x10^4 rad/sec
ω2= 5.42x10^4 rad/sec
ωmax= 8.42x10^4 rad/sec
θ= 1.72x10^4 rad
α=7.67 x10^4 rad/sec²
t= ωmax / α
t= 8.42 x10^4 rad/sec / 7.67 x10^4 rad/sec²
t=1.097 sec
