Isotopes are atoms with identical numbers of __ and different numbers of _

a satellite is traveling in a orbit around the earth. what change would cause it to experience more gravitational force from the

never [62]

The correct answer is:
<span>A. orbiting closer to the earths surface.

In fact, the gravitational force exerted by the Earth on the satellite is
</span> $F=G \frac{Mm}{r^2}$
<span>where
G is the gravitational constant
M is the Earth mass
m is the satellite mass
r is the distance of the satellite from the Earth's surface

We can see that, if the satellite orbits closer to the Earth's surface, its distance r from the centre of the planet decreases. But when r decreases, F (the gravitational force) increases, so A is the correct answer.</span>

7 0

2 years ago

Which layer of the atmosphere is directly above the troposhpere? A.troposhpere B.stratosphere C.mesosphere D.exoshpere

vivado [14]

Its B. i hope i helped!!!

4 0

3 years ago

Read 2 more answers

In terms of their location, what is the difference between subcutaneous fat and visceral fat

exis [7]

Subcutaneous fat is the fat that lies under the skin and isn’t necessarily harmful to your health and visceral fat is the harmful unseen fat surrounding the organs.

4 0

3 years ago

In a closed ring a and in an open ring magnet are falling along the x axis of thrme ring .the current generated in a and b have

zhenek [66]

Answer:

ALL AWNERS HERE

Explanation:

https://quizlet.com/449884025/test-3-physics-2-flash-cards/

8 0

2 years ago

Calculate the orbital period of a dwarf planet found to have a semimajor axis of a = 4.0x 10^12 meters in seconds and years.

padilas [110]

Explanation:

We have,

Semimajor axis is $4\times 10^{12}\ m$

It is required to find the orbital period of a dwarf planet. Let T is time period. The relation between the time period and the semi major axis is given by Kepler's third law. Its mathematical form is given by :

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

G is universal gravitational constant

M is solar mass

Plugging all the values,

$T^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.98\times 10^{30}}\times (4\times 10^{12})^3\\\\T=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.98\times10^{30}}\times(4\times10^{12})^{3}}\\\\T=4.37\times 10^9\ s$

Since,

$1\ s=3.17\times 10^{-8}\ \text{years}\\\\4.37\times 10^9\ s=4.37\cdot10^{9}\cdot3.17\cdot10^{-8}\\\\4.37\times 10^9\ s=138.52\ \text{years}$

So, the orbital period of a dwarf planet is 138.52 years.

3 0

2 years ago

Isotopes are atoms with identical numbers of __ and different numbers of __.

Isotopes are atoms with identical numbers of and different numbers of .