Ask question

Ask question

All categories

julia-pushkina [17]

3 years ago

15

Three identical satellites, X, Y, and Z are in outer space. The distance from satellite X to Y is 450 km, the distance from Y to

Z is 600 km, and the distance from satellite Z to X is 450 km. Between which satellites will the force of gravity be the greatest?
A. Between satellites Y and Z only.
B. Between satellites X and Y, and satellites Y and Z.
C. Between satellites Y and Z, and satellites X and Z.
D. Between satellites X and Y and satellites Z and X.

1 answer:

hodyreva [135]3 years ago

3 0

Gravity is stronger the closer you get. It is D.

You might be interested in

One particle has a charge of -1.87 x 10-9 C, while another particle has a charge of -1.10 x 10-9 C. If the two particles are sep

bogdanovich [222]

Answer:D 7.41 x 10-6 N

Explanation: AP*X

5 0

2 years ago

A ball is falling from the sky down to the left. What two forces are being used?

suter [353]

Answer:

push and pull

Explanation:

i beleve good luck❤️❤️

4 0

2 years ago

Read 2 more answers

A 2.0-kg ball rolls to the right at 3.0 m/s. A 4.0-kg ball rolls to the left at 2.0 m/s . What is the momentum of the system aft

charle [14.2K]

Answer:

Final momentum after a head on collision is -2kgm/s

Explanation:

One ball moves to the right and the other moves opposite and momentum is a vector quantity so that considering the direction

Initial momenta are P₁=2x3=6kgm/s P₂=4x(-2)=-8kgm/s

Final momentum is the vector sum of P(final)= 6-8= -2 kgm/s

4 0

3 years ago

What two energies does a bulldozer use

sesenic [268]

Mechanical and electrical

3 0

2 years ago

Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, m is m

netineya [11]

Answer: $m \frac{d}{dt}v_{(t)}$

Explanation:

In the image attached with this answer are shown the given options from which only one is correct.

The correct expression is:

$m \frac{d}{dt}v_{(t)}$

Because, if we derive velocity $v_{t}$ with respect to time $t$ we will have acceleration $a$ , hence:

$m \frac{d}{dt}v_{(t)}=m.a$

Where $m$ is the mass with units of kilograms ( $kg$ ) and $a$ with units of meter per square seconds $\frac{m}{s}^{2}$ , having as a result $kg\frac{m}{s}^{2}$

The other expressions are incorrect, let’s prove it:

$\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)}$ This result has units of $kg\frac{m}{s}$

$m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m$ This result has units of $kg$

$m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C$ This result has units of $kgm^{2}$ and $C$ is a constant

$m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m$ This result has units of $kg$

$m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m$ This result has units of $kg$

$\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C$ This result has units of $kg \frac{m^{3}}{s^{3}}$ and $C$ is a constant

$m\int a_{(t)} dt= \frac{m {a_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{4}}$ and $C$ is a constant

$\frac{m}{2} \frac{d}{dt}{(v_{(x)})}^{2}=0$ because $v_{(x)}$ is a constant in this derivation respect to $t$

$m\int v_{(t)} dt= \frac{m {v_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{2}}$ and $C$ is a constant

6 0

3 years ago