<h3><u>Question: </u></h3>
The equation for the speed of a satellite in a circular orbit around the Earth depends on mass. Which mass?
a. The mass of the sun
b. The mass of the satellite
c. The mass of the Earth
<h3><u>Answer:</u></h3>
The equation for the speed of a satellite orbiting in a circular path around the earth depends upon the mass of Earth.
Option c
<h3><u>
Explanation:
</u></h3>
Any particular body performing circular motion has a centripetal force in picture. In this case of a satellite revolving in a circular orbit around the earth, the necessary centripetal force is provided by the gravitational force between the satellite and earth. Hence 
.
Gravitational force between Earth and Satellite: 
Centripetal force of Satellite :
Where G = Gravitational Constant
= Mass of Earth
= Mass of satellite
R= Radius of satellite’s circular orbit
V = Speed of satellite
Equating 
, we get
Speed of Satellite 
Thus the speed of satellite depends only on the mass of Earth.
Answer: voltage, explode is the answers
Explanation:
Bc of the device for each server
Answer:
Option c. Inter-rater Reliability
Explanation:
Here, the rating is done by a group of data collectors under training for evaluation children's pain on Faces scale which is a scale ranging from 0 to 10 with different expressions or faces with a happy face at 0 to a crying face at 10.
Also in Inter-rater Reliability, the relative consistency of a study or test is assessed and the extent to which different group members rated the same behavior, the consistency of which is evaluated.
Thus it can also be helpful in interviews, etc.
Answer:
C) True. S increases with time, v₁ = gt and v₂ = g (t-t₀) we see that for the same t v₁> v₂
Explanation:
You have several statements and we must select which ones are correct. The best way to do this is to raise the problem.
Let's use the vertical launch equation. The positive sign because they indicate that the felt downward is taken as an opponent.
Stone 1
y₁ = v₀₁ t + ½ g t²
y₁ = 0 + ½ g t²
Rock2
It comes out a little later, let's say a second later, we can use the same stopwatch
t ’= (t-t₀)
y₂ = v₀₂ t ’+ ½ g t’²
y₂ = 0 + ½ g (t-t₀)²
y₂ = + ½ g (t-t₀)²
Let's calculate the distance between the two rocks, it should be clear that this equation is valid only for t> = to
S = y₁ -y₂
S = ½ g t²– ½ g (t-t₀)²
S = ½ g [t² - (t²- 2 t to + to²)]
S = ½ g (2 t t₀ - t₀²)
S = ½ g t₀ (2 t -t₀)
This is the separation of the two bodies as time passes, the amount outside the Parentheses is constant.
For t <to. The rock y has not left and the distance increases
For t> = to. the ratio (2t/to-1)> 1 therefore the distance increases as time
passes
Now we can analyze the different statements
A) false. The difference in height increases over time
B) False S increases
C) Certain s increases with time, v₁ = gt and V₂ = g (t-t₀) we see that for the same t v₁> v₂
