Question

suppose that throughout the united states, 350.0 x 10^6 suck braking processes occur in the course of a given day. calculate the average rate (megawatts) at which energy us being

Answer 1

Full Question:

Consider an automobile with a mass of 5510 lbm braking to a stop from a speed of 60.0 mph.

How much energy (Btu) is dissipated as heat by the friction of the braking process?

______Entry field with incorrect answer Btu

Suppose that throughout the United States, 350.0 x 10^6 such braking processes occur in the course of a given day.   Calculate the average rate (megawatts) at which energy is being dissipated by the resulting friction.

______Entry field with incorrect answer MW

Answer:

Energy dissipated as heat = 852.10 Btu

Average rate of energy dissipation = 3.64 MW

Explanation:

1 lb = 0.453592 Kg

Mass = 5510 lb = 0.453592 * 5510 Kg = 2499.29 Kg

1 mph = 0.44704 m/s

Velocity = 60 mph = 60 * 0.44704 m/s = 26.82 m/s

Let E be equal to energy acquired

$E = 1/2 MV^{2}$

$E = 0.5 * 2499.29 * 26.88^{2} \\E = 899.01 * 10^{3} Joules$

Energy dissipated as heat:

$E= 8.99*10^{3} = 0.0009478 * 8.99*10^{3}\\E = 852.10 tu$

E = 852.10 Btu

Energy dissipated throughout the United States

$E = 350 * 10^{6} * 899.01 *10^{3} \\E=3.1466 * 10^1^4 Joules$

Average power rate in MW, P

$P = (3.1466*10^{14} )/(24*60*60)\\P=3.64*10^{9} \\P=3.64MW$