Complete question is;
Shoveling snow can be extremely taxing since the arms have such a low efficiency in this activity. Suppose a person shoveling a sidewalk metabolizes food at the rate of 800 W. (The efficiency of a person shoveling is 3%.)
(a) What is her useful power output? (b) How long will it take her to lift 3000 kg of snow 1.20 m? (This could be the amount of heavy snow on 20 m of footpath.) (c) How much waste heat transfer in kilojoules will she generate in the process?
Answer:
A) P_out = 24 W
B) t = 1470 s
C) Q = 1140.72 KJ
Explanation:
We are given;
Input Power; P_in = 800 W
Efficiency; η = 3% = 0.03
A) Formula for efficiency is;
η = P_out/P_in
Making P_out the subject, we have;
P_out = η•P_in
P_out = 0.03 × 800
P_out = 24 W
B) We know that;
Power = work done/time taken
Thus;
P_out = mgh/t
We are given;
m = 3000 kg
h = 1.20 m
Thus, time is;
t = (3000 × 9.8 × 1.2)/24
t = 1470 s
C) amount of heat wasted is calculated from;
Q = (P_in - P_out)t
Q = (800 - 24) × 1470
Q = 1,140,720 J
Q = 1140.72 KJ
Answer: 4 herz is the answer!
Explanation:
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Answer:
Kinetic energy = (1/2) (mass) (speed²)
Original KE = (1/2) (1430 kg) (7.5 m/s)² = 40,218.75 joules
Final KE = (1/2) (1430 kg) (11.0 m/s)² = 86,515 joules
Work done during the acceleration = (40218.75 - 86515) = 46,296.25 joules
Power = work/time = 46,296.25 joules / 9.3 sec = 4,978.1 watts .
Explanation:
Dont report my answer please
As we know that KE and PE is same at a given position
so we will have as a function of position given as
also the PE is given as function of position as
now it is given that
KE = PE
now we will have
so the position is 0.707 times of amplitude when KE and PE will be same
Part b)
KE of SHO at x = A/3
we can use the formula
now to find the fraction of kinetic energy
now since total energy is sum of KE and PE
so fraction of PE at the same position will be
