Stays lined up with North Star
The work done on the car is -20 J.
Work done on the car is negative, meaning that the car actually does work on the external system.
<h3>Energy and law of conservation of energy</h3>
- Energy is the ability to do work
- the law of conservation of energy states that the total energy in a system is conserved
From the law of conservation of energy, the initial energy of the car before it moves down the road remains constant or unchanged.
- Initial energy = 100 J
- Initial energy = Final energy - work done on car
- Final Energy = Work done on car + initial energy
80J = Work done on car + 100 J
Work done on car = 80 - 100J
Work done on car = -20 J
Hence, the work done on the car is -20 J
Work done on car is negative.
Since work done on the car is negative, it means that the car actually does work on the external system. Hence, the decrease in the energy of the car.
Learn more about energy and work at: brainly.com/question/13387946
30000 btuh /3413 btuh/kw. = 8.8 kw
8.8 kw/.746 kw/hp = 11.8 hp if COP is 1
11.8/3 hp (COP coefficient of performance) = 3.99 COP
>>>So yes a 3.0 hp compressor with a nominal COP of 4 will handle the 30,000 btuh load.
3.2 to 4.5 is expected COP range for an air cooled heat pump or a/c unit.
A distance of d is covered with 53 mile/hr initially.
Time taken to cover this distance t1 = d/53 hour
Next distance of d is covered with x mile hours.
Time taken to cover this distance t2 = d/x hours.
We have average speed = 26.5 mile / hour
= Total distance traveled/ total time taken
= 
Answer:
The thrown rock strike 2.42 seconds earlier.
Explanation:
This is an uniformly accelerated motion problem, so in order to find the arrival time we will use the following formula:
So now we have an equation and unkown value.
for the thrown rock
for the dropped rock
solving both equation with the quadratic formula:
we have:
the thrown rock arrives on t=5.4 sec
the dropped rock arrives on t=7.82 sec
so the thrown rock arrives 2.42 seconds earlier (7.82-5.4=2.42)
