What is gravitational potential energy?

A 910-kg object is released from rest at an altitude of 1200 km above the north pole of the Earth. If we ignore atmospheric fric

gayaneshka [121]

In order to develop this problem it is necessary to use the concepts related to the conservation of both potential cinematic as gravitational energy,

$KE = \fract{1}{2}mv^2$

$PE = GMm(\frac{1}{r_1}-(\frac{1}{r_2}))$

Where,

M = Mass of Earth

m = Mass of Object

v = Velocity

r = Radius

G = Gravitational universal constant

Our values are given as,

$m = 910 Kg$

$r_1 = 1200 + 6371 km = 7571km$

$r_2 = 6371 km,$

Replacing we have,

$\frac{1}{2} mv^2 = -GMm(\frac{1}{r_1}-\frac{1}{r_2})$

$v^2 = -2GM(\frac{1}{r_1}-\frac{1}{r_2})$

$v^2 = -2*(6.673 *10^-11)(5.98 *10^24) (\frac{1}{(7.571 *10^6)} -\frac{1}{(6.371 *10^6)})$

$v = 4456 m/s$

Therefore the speed of the object when striking the surface of earth is 4456 m/s

3 0

4 years ago

At t = 0 the end you are oscillating is at its maximum positive displacement and is instantaneously at rest. Write an equation f

vovikov84 [41]

Answer:

The equation of displacement is $y=A\sin(\omega t-2.50 k+\dfrac{\pi}{2})$ .

Explanation:

Given that,

Distance = 2.50 m

We need to calculate the equation of wave

Using general equation of wave

$y=A\sin(\omega t-kx+\phi)$ ....(I)

Where, A = amplitude

t = time

x = displacement

$\phi$ = phase difference

Put the value in the equation

At t = 0, x = 0, y =A

$A=A\sin(0+\phi)$

$\sin\phi=1$

$\phi=\dfrac{\pi}{2}$

From equation (I)

$y=A\sin(\omega t-2.50 k+\dfrac{\pi}{2})$

Hence, The equation of displacement is $y=A\sin(\omega t-2.50 k+\dfrac{\pi}{2})$ .

7 0

3 years ago

Why is people asking for segges this is so annoying how do i block them help mee

NikAS [45]

Answer:

woah

Explanation:

report

6 0

3 years ago

PLS PLS HELP TIMED TEST!!!!!!

PolarNik [594]

Answer:

hi

Explanation:

hey

6 0

3 years ago

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A negative slope on the velocity vs. time graph indicates that the object is not

Svetlanka [38]

Answer: False, I think

Explanation:

Well it depends. Its hard to say without a function of the graph. Given a velocity graph, the first derivative is the acceleration.

Negative slope on a velocity graph means you are going in the negative direction. And over time, your velocity per unit of time is also decreasing. That means we are accelerating in the negative direction

4 0

3 years ago

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