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aleksley [76]
3 years ago
9

A battery-operated car utilizes a 12.0 V system. Find the charge the batteries must be able to move in order to accelerate the 7

50 kg car from rest to 25.0 m/s, make it climb a 2.00×102 m high hill, and then cause it to travel at a constant 25.0 m/s by exerting a 5.00×102 N force for an hour.
Physics
1 answer:
Yuliya22 [10]3 years ago
3 0

Answer:

3894531 coulombs

Explanation:

1 hour = 3600 seconds

Let g = 10m/s2

The distance that the car travel at 25 m/s over an hour is

s = 25 * 3600 = 90000 m

The total mechanical energy of the car is the sum of its kinetic energy to reach 25 m/s, its potential energy to climb up 200m high hill and it work to travel a distance of s = 90000m with F = 500 N force:

\sum E = E_k + E_p + E_W

\sum E = mv^2/2 + mgh + Fs

\sum E = 750*25^2/2 + 750*10*200 + 500*90000 = 46734375 J

This energy is drawn from the battery over an hour (3600 seconds), so its power must be

P = E / t = 46734375/3600 = 12982 W

The system is 12V so its current is

I = P/U = 12982 / 12 = 1081.8 A or 1081.8 Coulombs/s

The the total charge it needs for 1 hour (3600 s) is

C = 1081.8 * 3600 = 3894531 coulombs

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Please find the answer in the explanation

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An aircraft has a liftoff speed of 33 m/s. What is the minimum constant acceleration an
JulsSmile [24]

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a 10.0 kg sphere is released from rest in an ocean. as it falls, the water applies a resistive force r
dimaraw [331]

The calculated coefficient of kinetic friction is 0.33125.'

The rate of kinetic friction the friction force to normal force ratio experienced by a body moving on a dry, uneven surface is known as k. The friction coefficient is the ratio of the normal force pressing two surfaces together to the frictional force preventing motion between them. Typically, it is represented by the Greek letter mu (). In terms of math, is equal to F/N, where F stands for frictional force and N for normal force.

given mass of the block=10 kg

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now according to principal of conservation of energy we observe,

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The complete question is- A 10.0−kg block is released from rest at point A in Fig The track is frictionless except for the portion between point B and C, which has a length of 6.00m the block travels down the track, hits a spring of force constant 2250N/m, and compresses the spring 0.300m form its equilibrium position before coming to rest momentarily. Determine the coefficient of kinetic friction between the block and the rough surface between point Band (C)

Learn more about kinetic friction here-

brainly.com/question/13754413

#SPJ4

4 0
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