Answer:
Change in potential energy = 7350 Joules
Explanation:
It is given that,
Side of cube, a = 0.5 m
Density of cube, 
The cube is lifted vertically by a crane to a height of 3 m
We know that, density 
So, m = d × V (V = volume of cube = a³)

m = 250 kg
We have to find the change in potential energy of the cube. At ground level, the potential energy is equal to 0.
Potential energy at height h is given by :

PE = 250 kg × 9.8 m/s² ×3 m
PE = 7350 Joules
So, change in potential energy of the cube is 7350 Joules.
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.
Answer:
jfjcgufnfhfufm TV fifnricnrhkddufnfif km fgkfkvntfmrugrhfifnh r
Answer:
Explanation:
Using the formula for calculating range expressed as;
R = U√2H/g
U is the speed = 300m/s
H is the maximum height = 78.4m
g is the acceleration due to gravity = 9.8m/s²
Substitute into the fromula;
R = 300√2(78.4)/9.8
R = 300 √(16)
R = 300*4
R = 1200m
Hence the projectile travelled 1200m before hitting the ground
Answer:
go to the link quizzlet it will give you tha answer
Explanation: