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2 years ago

A cart is being pulled forward by a hanging mass. What is the value of the change in its kinetic energy, ΔEk?

1 answer:

8 0

The value of change in kinetic energy, ΔEk when a cart is being pulled forward by a hanging mass is Positive

In order words, ΔEk in the above question is positive

<h3>What is kinetic energy?</h3>

Kinetic energy can be defined as the energy a body possess when the body is in motion or due its motion

In conclusion, the value of change in kinetic energy, ΔEk when a cart is being pulled forward by a hanging mass is Positive

Learn more about kinetic energy:

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A uniform, solid, 1800.0 kgkg sphere has a radius of 5.00 mm. Find the gravitational force this sphere exerts on a 2.30 kgkg poi

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Answer

given,

Mass of the solid sphere = 1800 Kg

radius of the sphere,R = 5 m

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When Point is inside the Sphere

$F = \dfrac{GMm}{R^2}\ \dfrac{r}{R}$ when r<R

where r is the distance where the point mass is placed form the center

Now Force calculation

a) r = 5.05 m

$F = \dfrac{GMm}{r^2}$ [/tex]

$F = \dfrac{6.67\times 10^{-11}\times 1800\times 2.3}{5.05^2}$

F = 1.082 x 10⁻⁸ N

b) r = 2.65 m

$F = \dfrac{GMm}{R^2}\ \dfrac{r}{R}$

$F = \dfrac{6.67\times 10^{-11}\times 1800\times 2.3}{5.05^2}\ \dfrac{2.65}{5.05}$

$F = 5.68\times 10^{-9}\ N$

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You see a fish suspended in the water, based on this you can assume that:

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The weight of spaceman Speff at the surface of planet X, solely due to its gravitational pull, is 389 N. If he moves to a distan

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Answer

given,

W = weight of speff at surface of planet = 384 N

h= height = 1.86 x 10⁴ Km

W' = weight of speff at height h from the planet = 24.31 N

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M= mass of planet = ?

R = radius of planet

G = 6.67 x 10⁻¹¹ m³/kg.s²

$g = \dfrac{W}{M}$

$g = \dfrac{389}{75}$

g = 5.187 m/s²

we know,

$g = \dfrac{GM}{r^2}$

$5.19 = \dfrac{GM}{r^2}$ .......(1)

now,

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$g' = \dfrac{24.31}{75}$

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dividing equation (1)/(2)

$\dfrac{5.19}{0.324}=(\dfrac{(r+h)}{r})^2$

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