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enot [183]
3 years ago
8

The mass of the moon is 7.36×1022kg and its distance to the Earth is 3.84×108m. What is the gravitational force of the moon on t

he earth?
Physics
1 answer:
marissa [1.9K]3 years ago
8 0

Answer:

The gravitational force is F =  2\,*\,10^{20}\,N

Explanation:

To answer this question we need to recall Newton's Universal Law of Gravitation for the force "F" exerted from one object to the other:

F=G\,\frac{m_1\,*\,m_2}{d^2}

where G is the Universal gravitational constant = 6.674\,* \,10^{-11}\,\,\frac{m^3}{kg\,s^2}

m_1, and m_2 are the masses of the two bodies/objects attracting each other via gravitational force. In our case, one is the mass of the Earth = 5.972\,*\,10^{24}\, kg

and the other one,the mass of the Moon = 7.36\,*\,10^{22}\,kg

and lastly, "d" is the distance between to two objects. In our case:

d =3.84\,*\,10^8\,m

Since all these quantities are given in SI units, when we use them in the formula, our answer will result in the SI units of force "N" (Newtons):

F=6.674\,*10^{-11}\,\frac{5.972\,10^{24}\,7.36\,10^{22}}{(3.84\,10^8)^2} \,N\\F=1.989\,*\,10^{20}\,N\\

which can be rounded to: F = 2\,*\,10^{20}\,N

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Answer:

d) -4.0

Explanation:

The magnification of a lens is given by

M=-\frac{q}{p}

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M is the magnification

q is the distance of the image from the lens

p is the distance of the object from the lens

In this problem, we have

p = 50 cm is the distance of the object from the lens

q = 250 cm - 50 cm is the distance of the image from the lens (because the image is 250 cm from the obejct

Also, q is positive since the image is real

So, the magnification is

M=-\frac{200 cm}{50 cm}=-4.0

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A uniform disk of mass M = 4.9 kg has a radius of 0.12 m and is pivoted so that it rotates freely about its axis. A string wrapp
AlekseyPX

Answer:

(A) 2.4 N-m

(B) 0.035kgm^2

(C) 315.426 rad/sec

(D) 1741.13 J

(E) 725.481 rad

Explanation:

We have given mass of the disk m = 4.9 kg

Radius r = 0.12 m, that is distance = 0.12 m

Force F = 20 N

(a) Torque is equal to product of force and distance

So torque \tau =Fr, here F is force and r is distance

So \tau =20\times 0.12=2.4Nm

(B) Moment of inertia is equal to I=\frac{1}{2}mr^2

So I=\frac{1}{2}\times 4.9\times 0.12^2=0.035kgm^2

Torque is equal to \tau =I\alpha

So angular acceleration \alpha =\frac{\tau }{I}=\frac{2.4}{0.035}=68.571rad/sec^2

(C) As the disk starts from rest

So initial angular speed \omega _{0}=0rad/sec

Time t = 4.6 sec

From first equation of motion we know that \omega =\omega _0+\alpha t

So \omega =0+68.571\times  4.6=315.426rad/sec

(D) Kinetic energy is equal to KE=\frac{1}{2}I\omega ^2=\frac{1}{2}\times 0.035\times 315.426^2=1741.13J

(E) From second equation of motion

\Theta =\omega _0t+\frac{1}{2}\alpha t^2=0\times 4.6+\frac{1}{2}\times 68.571\times 4.6^2=725.481rad

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