All categories

3 years ago

a 1430 kg car speeds up from 7.50 m/s to 11.0 m/s in 9.30 s. Ignoring friction, how much power did that require?

1 answer:

iogann1982 [59]3 years ago

4 0

Kinetic energy = (1/2) (mass) (speed²)

Original KE = (1/2) (1430 kg) (7.5 m/s)² = 40,218.75 joules

Final KE = (1/2) (1430 kg) (11.0 m/s)² = 86,515 joules

Work done during the acceleration = (40218.75 - 86515) = 46,296.25 joules

Power = work/time = 46,296.25 joules / 9.3 sec = 4,978.1 watts .
_______________________________

One interesting thing to notice while you're (I'm) solving this problem
is the fact that although the car's speed only increased by about 47%,
its kinetic energy more than doubled ... increased by about 115%.
That's because kinetic energy is proportional to the SQUARE of the
speed.

You might be interested in

What will happen when these two waves interact?

oee [108]

B would be your answer

4 0

3 years ago

What is the most likely effect on the life of a plant if a student cuts it’s flowers?

goblinko [34]

Answer:

Explanation:

The reproduction parts of the plant are in the flower and around it so this would eliminate the plants ablity to reproduce.

5 0

3 years ago

Read 2 more answers

Biologists use optical tweezers to manipulate micron-sized objects using a beam of light. In this technique, a laser beam is foc

vekshin1

Answer:

Explanation:

Part A) Using

light intensity I= P/A

A= Area= π (Radius)^2= π((0.67*10^-6m)/(2))^2= 1.12*10^-13 m^2

Radius= Diameter/2

P= power= 10*10^-3=0.01 W

light intensity I= 0.01/(1.12*10^-13)= 9*10^10 W/m^2

Part B) Using

I=c*ε*E^2/2

rearrange to solve for E= $\sqrt{$ ((I*2)/(c*ε))

c is the speed of light which is 3*10^8 m/s^2

ε=permittivity of free space or dielectric constant= 8.85* 10^-12 F⋅m−1

I= the already solved light intensity= 8.85*10^10 W/m^2

amplitude of the electric field E= $\sqrt{$ (9*10^10 W/m^2)*(2) / (3*10^8 m/s^2)*(8.85* 10^-12 F⋅m−1)

---> E= $\sqrt{$ (1.8*10^11) / (2.66*10^-3) = $\sqrt{$ (6.8*10^13) = 8.25*10^6 V/m

8 0

3 years ago

A 1.50 3 103 - kg car starts from rest and accelerates uniformly to 18.0 m/s in 12.0 s. Assume that air resistance remains const

nexus9112 [7]

Answer:

Explanation:

let force exerted by engine be F.Net force =( F-400)N, applying newton law

F-400 = 1.5 x 10³x18 =27000 ,

F = 27400 N.

velocity after 12 s = 0 + 18 x 12 = 216 m/s

Average velocity = (0 + 216 )/2 = 108 m/s

Average power = force x average velocity = 27400 x 108 = 29.6 10⁵ W .⁶

b) At 12 s , velocity = 216 m/s

Instantaneous power = velocity x force = 216 x 27400 = 59.2 x 10⁶ W.

8 0

3 years ago

Read 2 more answers

A train increases its speed steadily from 10 m/s to 20 m/s in

rewona [7]

Answer:

15m/s

Explanation:

add the two speeds and divide by 2

10+20=30

30/2=15

3 0

3 years ago