Explain the role of gravity and inertia in keeping the moon in orbit? 15points pleaseee

Physics

1 answer:

lidiya [134]4 years ago

6 0

Answer:

gravity from the earth pulls the moon and the momentum of the rotaion and orbit of the moon keeps it moving

Explanation:

You might be interested in

To clearly see an image, the wavelength used must be at most 1/4 of the size of the object that is to be imaged. what frequency

LuckyWell [14K]

From the statement, the wavelength should be 1.6 cm * 1/4 = 0.4 cm. Then, we use the wave equation formula. The frequency is equal to the speed of light divided by the wavelength. First, let's use the SI units.

0.4 cm = 0.004 m

The speed of light is equal to 3 x 10^8 m/s

Frequency = 3 x 10^8 m/s ÷ 0.004 m
Frequency = 7.49 x 10^10 s^-1

5 0

3 years ago

Explain why radiation is dangerous to humans

tia_tia [17]

Radiation damages the cells that make up the human body, it can even cause cancer

7 0

4 years ago

Read 2 more answers

During a tennis match, a player serves the ball at 27.4 m/s, with the center of the ball leaving the racquet horizontally 2.34 m

Maurinko [17]

a. The ball's horizontal and vertical positions at time $t$ are given by

$x=\left(27.4\,\dfrac{\rm m}{\rm s}\right)t$

$y=2.34\,\mathrm m-\dfrac g2t^2$

The ball reaches the net when $x=12.0\,\rm m$ :

$12.0\,\mathrm m=\left(27.4\,\dfrac{\rm m}{\rm s}\right)t\implies t=0.438\,\rm s$

At this time, the ball is at an altitude of

$2.34\,\mathrm m-\dfrac g2\left(0.438\,\mathrm s\right)^2=1.40\,\mathrm m$

which is 1.40 m - 0.900 m = 0.500 m above the net.

b. The change in angle gives the ball the new position functions

$x=\left(27.4\,\dfrac{\rm m}{\rm s}\right)\cos(-5.00^\circ)t$

$y=2.34\,\mathrm m+\left(27.4\,\dfrac{\rm m}{\rm s}\right)\sin(-5.00^\circ)t-\dfrac g2t^2$

The ball reaches the net at time $t$ such that

$\left(27.4\,\dfrac{\rm m}{\rm s}\right)\cos(-5.00^\circ)t=12.0\,\mathrm m\implies t=0.440\,\mathrm s$

at which point the ball's vertical position would be

$2.34\,\mathrm m+\left(27.4\,\dfrac{\rm m}{\rm s}\right)\sin(-5.00^\circ)\left(0.440\,\mathrm s\right)-\dfrac g2\left(0.440\,\mathrm s\right)^2=0.343\,\mathrm m$