a 1.2x10^3 kilogram car is accelerated uniformly from 10. meters per second to 20 meters per second in 5.0 seconds. what is the

Response: a) 0.04 kW = 40 W

b) 0.05

Explanation:

A)

The thermal efficiency of the power cycle is calculated as Input / Output

Input = 10 kW + 14,400 kJ/min which translates to 10 kW + 14,400 kJ/(60s) = 10 kW + 14,400/60 kW.

Output equals 10 kW

Thus, Thermal Efficiency = Output / Input = 10 kW / 250 kW = 0.04 kW = 40 W

B)

Maximum Thermal Efficiency of the power cycle is defined as 1 - T1/T2

where T1 = 285 Kelvin

and T2 = 300 Kelvin

Thus, Maximum Thermal Efficiency = 1 - T1/T2 = 1 - 285/300 = 0.05

Answer:

P = 68.125 kW

P startup = 43.05 kW

Explanation:

Answer:

Explanation:

We start with the fact that

Initially, the spacecraft was at rest, u = 0

The final velocity of the rocket is given as v = 11 m/s

The distance that the rocket covers during acceleration is given as d = 213 m

We seek to determine the acceleration that the occupants of the spacecraft experience during launch, which is derived from the principles of kinematics. By applying the

third motion equation we can find the acceleration.

Thus, the acceleration felt by those inside the spacecraft is .