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Monica [59]
3 years ago
14

A 63 kg kg person starts traveling from rest down a waterslide 6.0 mm above the ground. At the bottom of the waterslide, it then

curves upwards by 1.0 mm above the ground such that the person is consequently launched into the air. Ignoring friction, how fast is the person moving upon leaving the waterslide
Physics
1 answer:
gladu [14]3 years ago
5 0

Answer:

change waterslide according to question. and you are good to go. check photo for solve

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A rock is stuck deep in the dirt. When you pull on the rock to remove it, what type of force are you exerting?
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SCALCET8 3.9.018.MI. A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the
Firlakuza [10]

Answer:

The length of his shadow is decreasing at a rate of 1.13 m/s

Explanation:

The ray of light hitting the ground forms a right angled triangle of height H, which is the height of the building and width, D which is the distance of the tip of the shadow from the building.

Also, the height of the man, h which is parallel to H forms a right-angled triangle of width, L which is the length of the shadow.

By similar triangles,

H/D = h/L

L = hD/H

Also, when the man is 4 m from the building, the length of his shadow is L = D - 4

So, D - 4 = hD/H

H(D - 4) = hD

H = hD/(D - 4)

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H = 2 m × 12 m/(12 m - 4 m)

H = 24 m²/8 m

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Since L = hD/H

and h and H are constant, differentiating L with respect to time, we have

dL/dt = d(hD/H)/dt

dL/dt = h(dD/dt)/H

Now dD/dt = velocity(speed) of man = -1.7 m/s ( negative since he is moving towards the building in the negative x - direction)

Since h = 2 m and H = 3 m,

dL/dt = h(dD/dt)/H

dL/dt = 2 m(-1.7 m/s)/3 m

dL/dt = -3.4/3 m/s

dL/dt = -1.13 m/s

So, the length of his shadow is decreasing at a rate of 1.13 m/s

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