please finish all 4 of the related subsections and keep the steps neat so I can

Question

please finish all 4 of the related subsections and keep the steps neat so I can understand what's going on. thank u so much for ur patience

of power, can produce the same level of brightness as a conventional lightbulb operating at power 100 w An . The lifetime of the energy- efficient bulb is 10 000 h and its purchase price is $4.60, whereas the conventional bulb has lifetime 750 h and costs $0.50 energy-efficient bulb over its lifetime as opposed to using conventional bulbs over the same time period Determine the total savings obtained by using one . Assume an energy cost of $0:20 per kilowatt-hour. 8 8 P034 An capable of supplying 140 Wh of energy. If friction forces and other losses account for 60.0% of the energy usage, what altitude change can a rider achieve when driving in hilly terrain, if the rider and scooter have a combined weight of 825 N? 1 points SerPSE8 8 P.035 When an automobile moves with constant speed down a highway, most of the power developed by the engine is used to compensate for the mechanical energy loss due to frictional forces exerted on the car by the air and the road. If the power developed by the engine is 177 hp, estimate the total friction force acting on the car when it is moving at a speed of 24 m/s. One horsepower equals 746 W. -11 points SerPSE8 & P.036 An older-model car accelerates from 0 to speed v in a time interval of At. A newer, more powerful sports car accelerates from O to Sv in the same time period. Assuming the energy coming from the engine appears only as kinetic energy of the cars, compare the power of the two cars Power older 35 PM

Explanation / Answer

Part A)

Energy consumed by energy efficent bulb for 10,000 hours is

E = 10000*(23/1000) = 230 KWh

Cost of energy efficent bulb for 10,000 hour including purchase price is

C = 230*0.2 + 4.6 = $ 50.6

For Conventional bulb

Number of of conventional bulbs used for 10,000 hour is

N = 10000/750 = 14 bulbs

Purchase cost of bulbs

Cp = 14*0.5 = $ 7.0

Energy consumed by conventional bulb is

E = 10000*(100/1000) = 1000

Cost of Conventional bulb for 10,000 hour including purchase price is

C = 1000*0.2 + 7.0 = $ 207.00

so savings are

savings = 207 - 50.6

savings = $ 156.4

Part B)

Available energy = total energy - energy loss due to losses

Available energy = 100%*E - 60%*E = 40%*E

E = 140 Wh = 140*3600 = 504000 J

40%*E = 201600 J

Available energy = 201600 J

Mechanical workdone = mgh

given, mg = 825N

Hence,

available energy = mechanical work done

201600 = 825* h

Altitude, h = 244.36 m

Part C)

given

power, P = 177 hp

= 177*746

P = 132042 watts

v = 24 m/s

let F is magnitide of frictional force acting on the car.

we know, P = F*v

F = P/v

= 132042/24

Total frictional force, F = 5501.75 N

Part D)

The power of two cars is given as the product of applied force and the speed.

Power = Force*speed

P = F*v

Polder= Folder. v

= {m*[(0 - v)/delta t]} . v

Polder = -mv2/delta t

Pnewer = Fnewer. v

= {m*[(0 - 5v)/delta t]} . 5v

Pnewer = -25mv2/delta t

Ratio, Pnewer / Polder = [-25mv2/delta t] / [-mv2/delta t]

Pnewer / Polder = 25

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