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4 years ago

13

A 15-kg ball is tossed up into the air. The ball is 2 meters off the ground traveling 4 m/s. What is the potential energy? A. 29

4 J B. 120 J C. 414 J D. 0

1 answer:

Sladkaya [172]4 years ago

7 0

Answer: 0j

Explanation:

At that point potential energy is zero and kinetic energy is maximum.. P. E=mgh=0

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The first and second velocities of the car are 0.25/t1 and 0.50/t2 respectively.

Given data:

The value of the first distance is, d1 = 0.25 m.

The value of the second distance is, d2 = 0.50 m.

The ratio of distance covered by an object in a specific direction and the time taken to cover the distance is known as the velocity of the object. Mathematically, the expression for the velocity is,

v = d/t

Here,

d is the distance covered.

t is the average time taken to cover the distance.

Then the first velocity of the car at 0.25 m is,

v1 = d1/t1

v1 = 0.25 / t1

here, t1 is the average time for first distance.

And the second velocity of the car with four washers at the 0. 50 m mark is,

v2 = d2/t2

v2 = 0.50 /t2

here, t2 is the average time for the second distance.

Thus, we can conclude that the first and second velocities of the car are 0.25/t1 and 0.50/t2 respectively.

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A motorcycle skids to a stop on a road.

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Friction of the road on the motorcycle in the opposite direction

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2 years ago

A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth. What multiple of Earth's radius RE g

7nadin3 [17]

Answer:

(a) r = 1.062·R $_E$ = $\frac{531}{500} R_E$

(b) r = $\frac{33}{25} R_E$

(c) Zero

Explanation:

Here we have escape velocity v $_e$ given by

$v_e =\sqrt{\frac{2GM}{R_E} }$ and the maximum height given by

$\frac{1}{2} v^2-\frac{GM}{R_E} = -\frac{GM}{r}$

Therefore, when the initial speed is 0.241v $_e$ we have

v = $0.241\times \sqrt{\frac{2GM}{R_E} }$ so that;

v² = $0.058081\times {\frac{2GM}{R_E} }$

v² = ${\frac{0.116162\times GM}{R_E} }$

$\frac{1}{2} v^2-\frac{GM}{R_E} = -\frac{GM}{r}$ is then

$\frac{1}{2} {\frac{0.116162\times GM}{R_E} }-\frac{GM}{R_E} = -\frac{GM}{r}$

Which gives

$-\frac{0.941919}{R_E} = -\frac{1}{r}$ or

r = 1.062·R $_E$

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$K_i = 0.241\times \frac{1}{2} \times m \times v_e^2 = 0.241\times \frac{1}{2} \times m \times \frac{2GM}{R_E} = \frac{0.241mGM}{R_E}$

Therefore we put $\frac{0.241GM}{R_E}$ in the maximum height equation to get

$\frac{0.241}{R_E} -\frac{1}{R_E} =-\frac{1}{r}$

From which we get

r = 1.32·R $_E$

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ME = KE - PE

Where the KE = PE required to leave the earth we have

ME = KE - KE = 0

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