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2 years ago

Write SHORT examples of Gravitational potential energy and Elastic energy

Physics

2 answers:

Nikolay [14]2 years ago

6 0

Dropping a bouncy ball and stretching a rubber ban.

gulaghasi [49]2 years ago

3 0

Answer:

Examples of a gravitational potential energy are; a flying helicopter, objects thrown up and examples of elastic energy are; an archers stretched bow, a stretched rubber band

Explanation:

We will start by giving a short definition of the two terminologies

Gravitational potential energy is the energy an object possess as a result of its position above the surface of the earth.

$E_{P}$ = mgh

where m = mass g = gravitational field strength h= height and $E_{P}$ =gravitational potential energy

Examples are; a flying helicopter, objects thrown up

Elastic energy: This occurs as a result of stretching, compressing or deforming of an object impermanently.

$E_{e}$ = $\frac{1}{2}$ ke²

where k = spring constant e=extension and $E_{e}$ =Elastic potential energy

examples of elastic energy are; an archers stretched bow, a stretched rubber band

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Karolina [17]

Answer:

276.62 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s² (positive downward and negative upward)

Equation of motion

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 3900+0^2}\\\Rightarrow v=276.62\ m/s$

<u>Neglecting air drag</u> the velocity of the spherical drop would be 276.62 m/s

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3 years ago

If you see lightening bolt and count for 4 seconds before you hear the thunder how far away was the lightening strike ? It was n

rjkz [21]

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3 years ago

Chegg Given that the mean radius of the Moon’s orbit is 3.84 x 108 m and its period is 2.36 x 106 sec, at what altitude above th

alukav5142 [94]

Answer:

The altitude of geostationary satellite is $3.58\times10^{7}\ m$

Explanation:

Given that,

Radius of moon's orbit $r=3.84\times10^{8}\ m$

Time period $T=2.36\times10^{6}\ sec$

We need to calculate the orbital radius of geostationary satellite is

Using formula of time period

$T=\sqrt{\dfrac{4\pi^2}{GM}a^3}$

$a=((\dfrac{GM}{4\pi^2})T^2)^{\dfrac{1}{3}}$

Where, G = gravitational constant

M = Mass of earth

T = time period of geostationary satellite orbit

Put the value in to the formula

$a=((\dfrac{6.67\times10^{-11}\times5.97\times10^{24}}{4\times\pi^2})\times(86160)^2)^{\dfrac{1}{3}}$

$a=4.217\times10^{7}\ m$

We need to calculate the altitude of geostationary satellite

Using formula of altitude

$h = a-R_{e}$

Where, R = radius of earth

a = radius of geostationary satellite

Put the value into the formula

$h =4.217\times10^{7}-6.38\times10^{6}$

$h =35790000\ m$

$h=3.58\times10^{7}\ m$

Hence, The altitude of geostationary satellite is $3.58\times10^{7}\ m$

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3 years ago

After a model rocket reached its maximum height, it then took 5.0 seconds to return to the launch site. what is the approximate

kifflom [539]

Air resistance is ignored.
g = 9.8 m/s².
At maximum height, the vertical velocity is zero.

Let h = the maximum height reached.
Let u = the vertical launch velocity.

Because ot takes 5.0 seconds to reach maximum height, therefore
(u m/s) - (9.8 m/s²)*(5 s) = 0
u = 49 m/s

The maximum height reached is
h = (49 m/s)*(5 s) - (1/2)*(9.8 m/s²)*(5 s)²
= 122.5 m

Answer: 122.5 m

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BaLLatris [955]

Answer:

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Explanation:

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