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natka813 [3]
3 years ago
8

A raccoon falls out of a tree from a height of 1.2. Which equation can you use to calculate the time it takes for the raccoon to

fall to the ground?
Physics
1 answer:
Inga [223]3 years ago
3 0

Answer:

t=\sqrt{\frac{2S}{g}}=\sqrt{\frac{2 \cdot 1.2 m}{9.81 m/s^2}}=0.50 s

Explanation:

The equation that we can use to calculate the time it takes for the raccoon to fall to the ground is:

t=\frac{2S}{g}

where S=1.2 m is the height of the tree and g=9.81 m/s^2 is the acceleration due to gravity. This equation is derived from the equation of the distance in a uniformly accelerated motion, which is given by

S=S_0 + v_0 t + \frac{1}{2}at^2

where S0 is the initial position, v0 is the initial velocity and t the time. In this problem, we can put S0=0 (we can take the initial position as the initial position of the raccoon) and v0=0 (the raccoon starts from rest), so the equation becomes

S=\frac{1}{2}at^2

and since the motion is a free fall, the acceleration is equal to the acceleration of gravity, so a=g:

S=\frac{1}{2}gt^2

And by re-arranging it, we find

t=\sqrt{\frac{2S}{g}}

By substituting numbers, we find

t=\sqrt{\frac{2S}{g}}=\sqrt{\frac{2 \cdot 1.2 m}{9.81 m/s^2}}=0.50 s

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Please HELP ME If the motion of an object changes, what must be true about the forces acting on that object?
atroni [7]

Answer:

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A motorcycle traveling at a speed of 44.0 mi/h needs a minimum of 44.0 ft to stop. If the same motorcycle is traveling 79.0 mi/h
Tasya [4]

Answer:

141.78 ft

Explanation:

When speed, u = 44mi/h, minimum stopping distance, s = 44 ft = 0.00833 mi.

Calculating the acceleration using one of Newton's equations of motion:

v^2 = u^2 + 2as\\\\v = 0 mi/h\\\\u = 44 mi/h\\\\s = 0.00833 mi\\\\=> 0^2 = 44^2 + 2 * a * 0.00833\\\\=> 1936 = -0.01666a\\\\a = -116206.48 mi/h^2 or -14.43 m/s^2

Note: The negative sign denotes deceleration.

When speed, v = 79mi/h, the acceleration is equal to when it is 44mi/h i.e. -116206.48 mi/h^2

Hence, we can find the minimum stopping distance using:

v^2 = u^2 + 2as\\\\v = 0 mi/h\\\\u = 79 mi/h\\\\a = -116206.48 mi/h\\\\=> 0^2 = 79^2 + (2 * -116206.48 * s)\\\\6241 = 232412.96s\\\\s = \frac{6241}{232412.96} \\\\s = 0.0268531 mi = 141.78 ft

The minimum stopping distance is 141.78 ft.

4 0
3 years ago
A helicopter, starting from rest, accelerates straight up from the roof of a hospital. The lifting force does work in raising th
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Answer:

25032.47 W

Explanation:

Power is the time rate of doing work, hence,

P = Work done(non conservative) / time

Work done (non conservative) is given as:

W = total K. E. + total P. E.

Total K. E. = 0.5mv²- 0.5mu²

Where v (final velocity) = 7.0m/s, u (initial velocity) = 0m/s

Total P. E. = mgh(f) - mgh(i)

Where h(f) (final height) = 7.2m, h(i) (initial height) = 0 m

=> W = 0.5mv² - mgh(f)

P = [0.5mv² - mgh(f)] / t

P = [(0.5*790*7²) - (790*9.8*7.2)] / 3

P = (19355 + 55742.4) / 3 = 75097.4/3

P = 25032.47 W

6 0
2 years ago
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