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myrzilka [38]
2 years ago
11

What is the net power needed to change the speed of a 1600-kg sport utility vehicle from 15.0 m/s to 40.0 m/s in 4.00 seconds

Physics
1 answer:
Sergio039 [100]2 years ago
3 0

Answer:

The net power needed to change the speed of the vehicle is 275,000 W

Explanation:

Given;

mass of the sport vehicle, m = 1600 kg

initial velocity of the vehicle, u = 15 m/s

final velocity of the vehicle, v = 40 m/s

time of motion, t = 4 s

The force needed to change the speed of the sport vehicle;

F = \frac{m(v-u)}{t} \\\\F = \frac{1600(40-15)}{4} \\\\F = 10,000 \ N

The net power needed to change the speed of the vehicle is calculated as;

P_{net} = \frac{1}{2} F[u + v]\\\\P_{net} = \frac{1}{2} \times 10,000[15 + 40]\\\\P_{net} = 275,000 \ W

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Explanation:

a)

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In this part A, we just need to analyze the horizontal motion. We know that:

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b)

For this part, we need to analyze the vertical motion of the projectile.

First, we want to find the initial vertical velocity. We can do it by using the equation for the vertical displacement:

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u_y is the initial vertical velocity

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We know that the total time of flight is t = 6.00 s: this means that when t=6, the projectile returns to its initial vertical position, so s = 0. Substituting and solving for u_y, we get

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To find the altitude of the highest point in the path, we use again the equation:

s=u_y t + \frac{1}{2}at^2

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Substituting the values, we find

s=(29.4)(3.00)+\frac{1}{2}(-9.8)(3.00)^2=44.1 m

So, the highest point is at 44.1 m above the ground.

Learn more about projectiles:

brainly.com/question/8751410

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