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
Well I know that ernest rutherford did the gold foil experiment where he fired alpha particles at gold foil. This experiment founded the nucleus but I don't know if the current model of the atom is based on this.
Cathode ray tube experiments sounds like its to do with electrolysis so i dont think it can be that.
<u>In modern physics</u>, as it was called "Stefan-Boltzmann law", the total energy radiated per unit surface area of a black body is directly proportional to the fourth power of the black body's temperature T
as:
where: P is the power (total energy radiated per second per square meter) and T is the temperature of a black body.
then we can make a ratio between the state of before quadruple (with subscript 1) and after (with subscript 2) as:
As
Then
then
- The factor will the total energy radiated per second per square meter increase = 256
