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3 years ago

How is work related to force and displacement?

Physics

1 answer:

VashaNatasha [74]3 years ago

3 0

Answer:

Displacement is the distance and direction an object is moved. Force and work are directly proportional to each other, while force and displacement are indirectly propotional. The equation showing the relationship is W= Fd. OR Work is a scalar. The rate at which work is done or at which energy is transferred. Work is equal to the force parallel to the displacement.

Explanation:

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pashok25 [27]

Here if we assume that there is no air friction on both balls then we can say

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4 years ago

Suppose that a planet were discovered between the sun and Mercury, with a circular orbit of radius equal to 2/3 of the average o

Musya8 [376]

Explanation:

It is given that,

A planet were discovered between the sun and Mercury, with a circular orbit of radius equal to 2/3 of the average orbit radius of Mercury.

Mass of the Sun, $M=1.99\times 10^{30}\ kg$

Radius of Mercury's orbit, $r=5.79\times 10^{10}\ m$

Radius of discovered planet, $R=\dfrac{2}{3}r$

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Let T is the orbital period of such a planet. Using Kepler's third law of planetary motion as :

$T^2\propto R^3$

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

$T^2=\dfrac{4\pi^2\times (3.86\times 10^{10})^3}{6.67\times 10^{-11}\times 1.99\times 10^{30}}$

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T = 4135214.625 s

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3 years ago

In 2004 astronomers reported the discovery of a large Jupiter-sized planet orbiting very close to the star HD 179949 (hence the

sp2606 [1]

<span>I'll tell you how to do it but you must crunch the numbers.

Use Kepler's 3rd Law

T^2 = k R^3

where k = 4(pi)^2/ GM
G =gravitational constant = 6.67300 × 10-11 m3 kg-1 s-2
M = mass of this new planet
pi = 3.14159265
T =3.09 days = 266976 seconds
R = (579,000,000km)/9 = 64333333.3 km

a)
Solve Kepler's 3rd Law for M. Your answer will be in kg

b)
mass of the sun = 1.98892 × 10^30 kilograms
Form the ratio
M(planet)/M(sun) </span>

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3 years ago

A supersonic jet is at an altitude of 14 kilometers flying at 1,500 kilometers per hour toward the east. At this velocity, how f

Kaylis [27]

Answer:

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$Distance=Speed\times Time$

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