Answer:t
Explanation:
Given that,
Power output is
P = 0.50kW = 0.5 × 1000 = 500W.
We want to find energy transfer in
t = 2minutes = 2 × 60 = 120 seconds
Also, time to hour
t = 2/60 = 1/30 hour
Power is the rate of energy.
Therefore,
Power = energy / time taken
Energy = power × time taken
Energy = 0.5kW × 1/30 hr
Energy = 0.01667 kWh
Also, energy in Joule
Energy = power × time
Energy = 500W × 120s
Energy = 60,000J
Energy = 60kJ.
Answer:
<h2>Because water can lose and accept H + ions .......</h2>
Answer:
The magnitude of the new electric field is <u>35820 N/C</u>.
Explanation:
Given:
Original magnitude of electric field (E₀) = 2388 N/C
Original voltage = 'V' (Assume)
Original separation between the plates = 'd' (Assume)
Now, new voltage is three times original voltage. So, 
New distance is 1/5 the original distance. So, 
Now, electric field between the parallel plates originally is given as:
Let us find the new electric field based on the above formula.
Now, 
. So,
Therefore, the magnitude of the new electric field is 35820 N/C.
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
