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cestrela7 [59]
3 years ago
8

A 2000 kg car accelerates from 0 to 25 m / s in 21.0 s. How much is the average power delivered by the motor? (1 hp = 746 W)

Physics
2 answers:
mash [69]3 years ago
3 0

Answer: 40.2 hp

Explanation:

The mass of the car is 2000kg.

in 21 seconds, it accelerates from 0m/s to 25m/s, so the average acceleration is:

a = (25m/s)/(21s) = 1.2 m/s^2

We know that force = mass*acceleration, then:

F = 2000kg*1.2m/s^2 = 2400N.

Now, the Power can be written as:

P = W/t where W is work = F*d (force per distance) and t is time.

then we have:

P = (F*d)/t and d/t  is the average velocity, and we know the velocity v = 25m/s, so the average velocity will be (1/2)*25m/s = 12.5m/s

P = F*12.5m/s

and the force is F = 2400N

P = 2400 N*12.5m/s = 30,000 W

and we know that 1hp = 746 W

then, P = (30,000/746) hp = 40.2 hp

DIA [1.3K]3 years ago
3 0

Answer:

P=40hp

Explanation:

Hello,

In this cased, for the given velocities, we can compute the work done by the car as shown below, considering the kinetic energy only:

W=\frac{1}{2}m(v_f^2-v_0^2)

Whereas the mass is 2000 kg, the final velocity is 25 m/s and the initial velocity is 0 m/s, therefore the work is:

W=\frac{1}{2}*2000kg*[(25\frac{m}{s})^2  -(0\frac{m}{s})^2]\\\\W=\frac{1}{2}*2000kg*625\frac{m^2}{s^2} \\\\W=6.25x10^5J

Now, the power:

P=\frac{W}{t}=\frac{6.25x10^5}{21.0s}=29762W

And in horsepower:

P=29762W*\frac{1hp}{746W} \\\\P=40hp

Best regards.

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A rock is thrown upward from the top of a 30 m building with a velocity of 5 m/s. Determine its velocity (a) When it falls back
castortr0y [4]

Answer:

a) 5 m/s downwards

b) 17.86 m/s

c) 24.82 m/s

d) 0.228

Explanation:

We can set the frame of reference with the origin on the top of the building and the X axis pointing down.

The rock will be subject to the acceleration of gravity. We can use the equation for position under constant acceleration and speed under constant acceleration:

X(t) = X0 + V0 * t + 1/2 * a * t^2

V(t) = V0 + a * t

In this case

X0 = 0

V0 = -5 m/s

a = 9.81 m/s^2

To know the speed it will have when it falls back past the original point we need to know when it will do it. When it does X will be 0.

0 = -5 * t + 1/2 * 9.81 * t^2

0 = t * (-5 + 4.9 * t)

One of the solutions is t = 0, but this is when the rock was thrown.

0 = -5 + 4.69 * t

4.9 * t = 5

t = 5 / 4.9

t = 1.02 s

Replacing this in the speed equation:

V(1.02) = -5 + 9.81 * 1.02 = 5 m/s (this is speed downwards because the X axis points down)

When the rock is at 15 m above the street it is 15 m under the top of the building.

15 = -5 * t + 1/2 * 9.81 * t^2

4.9 * t^s -5 * t - 15 = 0

Solving electronically:

t = 2.33 s

At that time the speed will be:

V(2.33) = -5 + 9.81 * 2.33 = 17.86 m/s

When the rock is about to reach the ground it is at 30 m under the top of the building:

30 = -5 * t + 1/2 * 9.81 * t^2

4.9 * t^s -5 * t - 30 = 0

Solving electronically:

t = 3.04 s

At this time it has a speed of:

V(3.04) = -5 + 9.81 * 3.04 = 24.82 m/s

---------------------

Power is work done per unit of time.

The work in this case is:

L = Ff * d

With Ff being the friction force, this is related to weight

Ff = μ * m * g

μ: is the coefficient of friction

L = μ * m * g * d

P = L/Δt

P = (μ * m * g * d)/Δt

Rearranging:

μ = (P * Δt) / (m * g * d)

1 horsepower is 746 W

20 minutes is 1200 s

μ = (746 * 1200) / (100 * 9.81 * 4000) = 0.228

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Answer:

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