How much potential energy does a 20-kg barrel have when it is in the truck whose floor is 5.0 m above the ground? P.E. = J

N which order did the events forming our solar system occur?

brilliants [131]

Answer:

The solar nebula became hot and dense because of that it pulling in more gas. This flattened into a rotating disk. It spun faster and faster, forming the Sun.

Explanation:

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4 0

3 years ago

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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

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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

If you wanted to detect x rays coming from the sun, where would you place the detector? Why?

Damm [24]

You would have to place your sensor above earth's atmosphere because it blocks out nearly all x-rays. this is why we have the Chandra observatory

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3 years ago

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In which of the basic regions of the galaxy is the sun located?

AlladinOne [14]

galactic disk

The galactic disk is a thinned, leveled out distribution of stars which includes the typical to the largest and brightest. The Sun is in the Milky Way and lies amongst the majority of the stars where it bulges.

8 0

4 years ago

A 3.00 kg mud ball has a perfectly inelastic collision with a second mud ball that is initially at rest. the composite system mo

viva [34]

Perfectly inelastic collision is type of collision during which two objects collide, stay connected and momentum is conserved. Formula used for conservation of momentum is:

$m_{1} * v_{1} + m_{2} * v_{2} = m_{1} * v'_{1} + m_{2} * v'_{2}$

In case of perfectly inelastic collision v'1 and v'2 are same.

We have following information:
m₁=3 kg
m₂=? kg
v₁=x m/s
v₂=0 m/s
v'1 = v'2 = 1/3 * v₁

Now we insert given information and solve for m₂:
3*v₁ + 0*? = 3*1/3*v₁ + m₂*1/3*v₁
3v₁ = v₁ + m₂*1/3*v₁
2v₁ = m₂*1/3*v₁
2 = m₂*1/3
m₂= 6kg

Mass of second mud ball is 6kg.

7 0

3 years ago