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babymother [125]
3 years ago
8

A bicyclist is traveling at a speed of 20.0 m/s as he approaches the bottom of a hill. He decides to coast up the hill and stops

upon reaching the top. Determine the vertical height of the hill. Ignore friction and air resistance.A) 28.5 m B) 3.70 m C) 11.2 m D) 40.8 m E) 20.4 m
Physics
1 answer:
Rudiy273 years ago
5 0

Answer:

Vertical height of the hill, h = 20.4 meters

Explanation:

Given that,

Speed of the bicyclist, v = 20 m/s

To find,

The vertical height of the hill.

Solution,

Let h is the height of the hill. When he approaches the bottom of the hill, the loss in kinetic energy is equal to the gain in potential energy. Using the conservation of energy as,

\dfrac{1}{2}mv^2=mgh

h=\dfrac{v^2}{2g}

h=\dfrac{(20)^2}{2\times 9.8}

h = 20.4 meters

Therefore, the vertical height of the hill is 20.4 meters. Hence, this is the required solution.

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A ball rolls for 8 seconds and travels 24 meters. How fast was it traveling?
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Answer:

The speed of the ball was, v = 3 m/s

Explanation:

Given data,

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The distance the ball rolled, d = 24 m

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8 0
3 years ago
1-A car moves toward east 12km is represented as A and it turns towards south 16km is represented as B. What is the resultant ve
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1. A-20 km south east

The car's displacement consists of two components into two different directions. Using a system of coordinates in which x represents the east direction and y represents the south direction, the two displacements are:

d_x = 12 km east

d_y = 16 km south

Since the two components are orthogonal to each other, we can find the resultant displacement by using Pythagorean's theorem:

d=\sqrt{d_x^2+d_y^2}=\sqrt{(12 km)^2+(16 km)^2}=\sqrt{400}=20 km

and the direction is between the two original directions, so south-east.

2. D. 10 m/s

First of all, we need to calculate the total time the stone took to hit the ground. Since the vertical distance covered is S = 78.4 m, and since the motion is an accelerated motion with constant acceleration g=9.8 m/s^2, we have

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

From which we find the total time of the fall, t:

t=\sqrt{\frac{2S}{g}}=\sqrt{\frac{2(78.4 m)}{9.8 m/s^2}}=4 s

Now we can consider the horizontal motion of the stone: we know that the stone travels for d = 40 m in a time of t = 4 s, therefore the horizontal velocity of the stone is

v=\frac{d}{t}=\frac{40 m}{4 s}=10 m/s

3. B=32.32 m

As in the previous problem, we have to calculate the total time it takes for the stone to reach the river first. Since the vertical distance covered is S = 20 m, we have

t=\sqrt{\frac{2S}{g}}=\sqrt{\frac{2(20 m)}{9.8 m/s^2}}=2.0 s

And since the stone is traveling horizontally at v = 16 m/s, the horizontal distance covered is

d=vt=(16 m/s)(2 s)=32 m

So, the closest answer is B.

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Genrish500 [490]

Answer:

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