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alexdok [17]
3 years ago
8

A 1.2 x 103 kg racecar, with a velocity of 8 m/s, collides with an unsuspecting 80 kg honey badger who is standing

Physics
1 answer:
aalyn [17]3 years ago
4 0

Answer: 90 m/s

Explanation:

Given

mass of racecar M=1.2\times10^3\ kg

velocity of racecar u=8\ m/s

mass of still honeybadger m=80\ kg

after collision race car is traveling at a speed of v_1=2\ m/s

conserving linear momentum

Mu+m\times0=Mv_1+ mv_2\quad[v_2=\text{velocity of honeybadger after colllision}]

1.2\times10^3\times8+0=1.2\times10^3\times2+80\times v_2

1.2\times10^3(8-2)=80v_2\\v_2=\frac{7.2\times10^3}{80}=90\ m/s

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Consider a satellite in a circular orbit around the earth. Why is it important to give a satellite a horizontal speed when placi
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4 years ago
Latitude and longitude are the two coordinates that determine a specific point on the Earth’s surface. How does knowing the loca
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hey sandra! i hope this will help you!

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6 0
3 years ago
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

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t = 5 / 4.9

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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

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t = 2.33 s

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

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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

8 0
3 years ago
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