m = Mass of the refrigerator to be moved to third floor = 136 kg
g = Acceleration due to gravity by earth on the refrigerator being moved = 9.8 m/s²
h = Height to which the refrigerator is moved = 8 m
W = Work done in lifting the object
Work done in lifting the object is same as the gravitational potential energy gained by the refrigerator. hence
Work done = Gravitation potential energy of refrigerator
W = m g h
inserting the values
W = (136) (9.8) (8)
W = 10662.4 J
Answer:
Temperature of water leaving the radiator = 160°F
Explanation:
Heat released = (ṁcΔT)
Heat released = 20000 btu/hr = 5861.42 W
ṁ = mass flowrate = density × volumetric flow rate
Volumetric flowrate = 2 gallons/min = 0.000126 m³/s; density of water = 1000 kg/m³
ṁ = 1000 × 0.000126 = 0.126 kg/s
c = specific heat capacity for water = 4200 J/kg.K
H = ṁcΔT = 5861.42
ΔT = 5861.42/(0.126 × 4200) = 11.08 K = 11.08°C
And in change in temperature terms,
10°C= 18°F
11.08°C = 11.08 × 18/10 = 20°F
ΔT = T₁ - T₂
20 = 180 - T₂
T₂ = 160°F
Answer:
0.34 m
Explanation:
From the question,
v = λf................ Equation 1
Where v = speed of sound, f = frequency, λ = Wave length
Make λ the subject of the equation
λ = v/f............... Equation 2
Given: v = 340 m/s, f = 500 Hz.
Substitute these values into equation 2
λ = 340/500
λ = 0.68 m
But, the distance between a point of rarefaction and the next compression point, in the resulting sound is half wave length
Therefore,
λ/2 = 0.68/2
λ/2 = 0.34 m
Hence, the distance between a point of rarefaction and the next compression point, in the resulting sound is 0.34 m
