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Ludmilka [50]
2 years ago
15

If a cheetah goes from 10 m/s to 30 m/s in 5 seconds, what was her acceleration?

Physics
2 answers:
grandymaker [24]2 years ago
5 0

Explanation:

Y CHETOS XDXDXX

CDXCVHH

Ierofanga [76]2 years ago
3 0
Yea i i didn’t see her last name but yet bbbsnsnsnsnddx i was talking abt it to see if he could see her what she was going on with lol but i yea i she said she was like he didn’t do that too but he said it ain’t no big do but it was a weird thing
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Help please!!!
TiliK225 [7]

Answer:

Cu 8.92

Explanation:

The formula for density is mass/volume. If you were to divide 62.44 by 7, you would get 8.92. Since copper is the only metal in this table that has a density of 8.92, that is the answer.

7 0
3 years ago
Determine the speed of a falling object 7.8 sec after it dropped.
butalik [34]

Answer:

76.4m/s

Explanation:

Given parameter:

Time taken  = 7.8sec

Unknown:

Speed after it dropped  = ?

Solution:

To solve this problem, we use one of the kinematics equation:

      V  = U + gt

V is the final speed

U is the initial speed  = 0m/s

g is the acceleration due to gravity

t is the time taken

       V  = 0 + 9.8 x 7.8  = 76.4m/s

7 0
2 years ago
Which of these scenarios describes circular motion?
Anika [276]

A)

The moon orbiting the Earth

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

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

8 0
2 years ago
The most frequent compulsion that is exhibited in obsessive-compulsive disorder is
Vinil7 [7]
The most frequent compulsion that is exhibited in obsessive compulsive disorder is cleansing 
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