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algol [13]
3 years ago
11

In a water park, people walk a distance of 20 meters up an inclined ramp before getting on a water slide. If the spot from where

they start sliding down is located 7 meters above the water’s surface, calculate the angle of the ramp with the horizontal.
Physics
2 answers:
Elenna [48]3 years ago
8 0
Thank you for posting your Physics question here. I hope the answer helps.  Upon calculating the ramp with the horizontal the answer is 20.49 Deg. Below is the solution:

Y = 7 m. 
<span>r = 20 m. </span>

<span>sinA = Y/r = 7/20 = 0.35. </span>
<span>A = 20.49 Deg.</span>
Elena L [17]3 years ago
7 0

Answer:

E.  20°

Explanation:

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Given,

Distance from the surface to the center of the earth, d=4000 miles

Distance from the center to you at a height of 8000 miles, a= 8000+4000=12000 miles

The gravitational force acting on a person at the surface is equal to his weight.

From Newton's Universal Law of Gravitation, the gravitational force is

F=\frac{G\times M\times m}{r^2}

Where G is the gravitational constant, M is the mass of the earth, m is the mass of the object/person, r is the distance between the center of the earth and the object/person

At the surface, this force is equal to the weight of the person, W=mg

i.e.

F_s=\frac{G\times M\times m}{d^2}=W

On substituting the of d,

W=\frac{\text{GMm}}{4000^2}

At a height of 8000 miles from the surface, the gravitational force is equal to,

F_a=\frac{GMm}{12000^2}

On dividing the above two equations,

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Therefore,

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Therefore at a height of 8000 miles above the surface of the earth, the force of gravity becomes 1/9 time your weight.

5 0
1 year ago
A car starts from rest and travels for t1 seconds with a uniform acceleration a1. The driver then applies the brakes, causing a
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A parallel-plate capacitor with plates of area 360 cm2 is charged to a potential difference V and is then disconnected from the
Softa [21]

Answer:

Q=3.9825\times 10^{-9} C

Explanation:

We are given that a parallel- plate capacitor is charged to a potential difference V and then disconnected from the voltage source.

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We have to find the charge Q on the positive plates of the capacitor.

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C=\frac{\epsilon_0 S}{d}

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C_1=\frac{\epsilon_0 S}{(d+\Delta d)}

V=\frac{Q}{C}

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V=\frac{Q}{C_1}

V+\Delta V=\frac{Q}{C_1}

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