V = IR
I = V/R
I = 12/6
I = 2 amps
The power that heat pump draws when running will be 6.55 kj/kg
A heat pump is a device that uses the refrigeration cycle to transfer thermal energy from the outside to heat a building (or a portion of a structure).
Given a heat pump used to heat a house runs about one-third of the time. The house is losing heat at an average rate of 22,000 kJ/h and if the COP of the heat pump is 2.8
We have to determine the power the heat pump draws when running.
To solve this question we have to assume that the heat pump is at steady state
Let,
Q₁ = 22000 kj/kg
COP = 2.8
Since heat pump used to heat a house runs about one-third of the time.
So,
Q₁ = 3(22000) = 66000 kj/kg
We known the formula for cop of heat pump which is as follow:
COP = Q₁/ω
2.8 = 66000 / ω
ω = 66000 / 2.8
ω = 6.66 kj/kg
Hence the power that heat pump draws when running will be 6.55 kj/kg
Learn more about heat pump here :
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Answer:
The value is 
Explanation:
From the question we are told that
The average speed for the first 12 hours is 
The average speed for the next 13 hours is 
Generally the total time taken is mathematically represented as

=> 
The distance covered in the first movement is



The distance covered in the first movement is



The total distance traveled is



The average of the whole journey is




Answer:
True
Explanation:
Moving the point charge away from the source increases the distance of separation.

from the above equation the potential V and r are inversely proportion so, it will surely decrease with an increase in distance.