How long does it take to run 150 meters at the speed of 18 km

During a solar eclipse, the Moon, Earth, and Sun all lie on the same line, with the Moon between the Earth and the Sun.

lord [1]

Answer:

(a) $F_{sm} = 4.327\times 10^{20}\ N$

(b) $F_{em} = 1.983\times 10^{20}\ N$

(c) $F_{se} = 3.521\times 10^{20}\ N$

Solution:

As per the question:

Mass of Earth, $M_{e} = 5.972\times 10^{24}\ kg$

Mass of Moon, $M_{m} = 7.34\times 10^{22}\ kg$

Mass of Sun, $M_{s} = 1.989\times 10^{30}\ kg$

Distance between the earth and the moon, $R_{em} = 3.84\times 10^{8}\ m$

Distance between the earth and the sun, $R_{es} = 1.5\times 10^{11}\ m$

Distance between the sun and the moon, $R_{sm} = 1.5\times 10^{11}\ m$

Now,

We know that the gravitational force between two bodies of mass m and m' separated by a distance 'r' is given y:

$F_{G} = \frac{Gmm'_{2}}{r^{2}}$ (1)

Now,

(a) The force exerted by the Sun on the Moon is given by eqn (1):

$F_{sm} = \frac{GM_{s}M_{m}}{R_{sm}^{2}}$

$F_{sm} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 7.34\times 10^{22}}{(1.5\times 10^{11})^{2}}$

$F_{sm} = 4.327\times 10^{20}\ N$

(b) The force exerted by the Earth on the Moon is given by eqn (1):

$F_{em} = \frac{GM_{s}M_{m}}{R_{em}^{2}}$

$F_{em} = \frac{6.67\times 10^{- 11}\times 5.972\times 10^{24}\times 7.34\times 10^{22}}{(3.84\times 10^{8})^{2}}$

$F_{em} = 1.983\times 10^{20}\ N$

(c) The force exerted by the Sun on the Earth is given by eqn (1):

$F_{se} = \frac{GM_{s}M_{m}}{R_{es}^{2}}$

$F_{se} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 5.972\times 10^{24}}{((1.5\times 10^{11}))^{2}}$

$F_{se} = 3.521\times 10^{20}\ N$

7 0

3 years ago

Calculate the electrical energy expended in a device across which the circuit voltage drops by 50 volts in moving a charge of 6

solmaris [256]

1 volt = 1 joule per coulomb 50 volts = 50 joules per coulomb 50 joules/coulomb times 6 coulombs = 300 joules

5 0

4 years ago

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When the forces acting on a particle are resolved into cylindrical components, friction forces always act in the

Aleksandr [31]

Answer:

Tangential

Explanation: This is a kind of force which act on a moving body in such a way that it is curved in the direction of the path of the body. This implies that when the velocity of the object is positive, the acceleration will be negative.

7 0

3 years ago

If the energy input of an engine is 6,000 J and its efficiency is 33%, calculate its energy output. Eout = ______ J 200 2000 300

AnnyKZ [126]

The answer to the question is 2,000 J. Hope this helps!

6 0

4 years ago

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What is the slit spacing of a diffraction necessary for a 600 nm light to have a first order principal maximum at 25.0°?

Sauron [17]

Explanation:

Given that,

Wavelength of light, $\lambda=600\ nm=6\times 10^{-7}\ m$

Angle, $\theta=25^{\circ}$

We need to find the slit spacing for diffraction. For a diffraction, the first order principal maximum is given by :

$d\sin\theta=n\lambda$

n is 1 here

d is slit spacing

$d=\dfrac{\lambda}{\sin\theta}\\\\d=\dfrac{6\times 10^{-7}}{\sin(25)}\\\\d=1.41\times 10^{-6}\ m\\\\d=1.41\ \mu m$

So, the slit spacing is $1.41\ \mu m$ .

6 0

4 years ago