4.0 is gonna be your answer
Answer:
The increase in potential energy of the ball is 115.82 J
Explanation:
Conceptual analysis
Potential Energy (U) is the energy of a body located at a certain height (h) above the ground and is calculated as follows:
U = m × g × h
U: Potential Energy in Joules (J)
m: mass in kg
g: acceleration due to gravity in m/s²
h: height in m
Equivalences
1 kg = 1000 g
1 ft = 0.3048 m
1 N = 1 (kg×m)/s²
1 J = N × m
Known data
Problem development
ΔU: Potential energy change
ΔU = U₂ - U₁
U₂ - U₁ = mₓgₓh₂ - mₓgₓh₁
U₂ - U₁ = mₓg(h₂ - h₁)
The increase in potential energy of the ball is 115.82 J
Answer:
0.358Kg
Explanation:
The potential energy in the spring at full compression = the initial kinetic energy of the bullet/block system
0.5Ke^2 = 0.5Mv^2
0.5(205)(0.35)^2 = 12.56 J = 0.5(M + 0.0115)v^2
Using conservation of momentum between the bullet and the block
0.0115(265) = (M + 0.0115)v
3.0475 = (M + 0.0115)v
v = 3.0475/(M + 0.0115)
plugging into Energy equation
12.56 = 0.5(M + 0.0115)(3.0475)^2/(M + 0.0115)^2
12.56 = 0.5 × 3.0475^2 / ( M + 0.0115 )
12.56 = 0.5 × 9.2872/ M + 0.0115
12.56 = 4.6436/ M + 0.0115
12.56 ( M + 0.0115 ) = 4.6436
12.56M + 0.1444 = 4.6436
12.56M = 4.6436 - 0.1444
12.56 M = 4.4992
M = 4.4992÷12.56
M = 0.358 Kg
Answer:
a) 2.85 kW
b) $ 432
c) $ 76.95
Explanation:
Average price of electricity = 1 $/40 MJ
Q = 20 kW
Heat energy production = 20.0 KJ/s
Coefficient of performance, K = 7
also
K=(QH)/Win
Now,
Coefficient of Performance, K = (QH)/Win = (QH)/P(in) = 20/P(in) = 7
where
P(in) is the input power
Thus,
P(in) = 20/7 = 2.85 kW
b) Cost = Energy consumed × charges
Cost = ($1/40000kWh) × (16kW × 300 × 3600s)
cost = $ 432
c) cost = (1$/40000kWh) × (2.85 kW × 200 × 3600s) = $76.95
