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:
Explanation:
From the Question we are told that:
Mass 
Coefficient of kinetic friction 
Generally the equation for Frictional force is mathematically given by
Generally the Newton's equation for Acceleration due to Friction force is mathematically given by
Therefore
Answer:
91.017N
Explanation:
Parameters
L=4.67m, m=0.192kg, t = 0.794s, The pulse makes four trips down and back along the cord, we have 4 +4 =8 trips( to and fro)
so N= no of trips = 8, From Wave speed(V) = N *L/t , we have :
V= 8*4.67/0.794 = 47.0529 m/s.
We compute the cords mass per length, Let it be P
P = M/L = 0.192/4.67 = 0.04111 kg/m
From T = P * V^2 where T = Tension, we have
T = 0.04111 * (47.0529)^2
T = 91.017N.
The tension in the cord is 91.017N
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