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tatyana61 [14]
3 years ago
8

The moon has been orbiting the earth at the same speed for millions of years. This is an example of Newton’s 1st Law because

Physics
1 answer:
alekssr [168]3 years ago
4 0

Answer:

an object remains at rest (or motion) unless acted upon by an outside force

Explanation:

force that created the moon imparted velocity which was restrained by the gravitational pull of the earth

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Hipparcos estimated its distance at roughly 96 parsecs from Earth, or 310 ± 20 light years away.
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Relay of thermostate is not working why
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7 0
3 years ago
If earth's mass were half its actual value but its radius stayed the same, the escape velocity of earth would be:________
siniylev [52]

If the earth's mass were half its actual value but its radius stayed the same, the escape velocity of the earth would be V_e = \sqrt{\dfrac{GM}{r}}.

<h3>What is an escape velocity?</h3>

The ratio of the object's travel distance over a specific period of time is known as its velocity. As a vector quantity, the velocity requires both the magnitude and the direction. the slowest possible speed at which a body can break out of the gravitational pull of a certain planet or another object.

The formula to calculate the escape velocity of earth is given below:-

V_e=\sqrt{\dfrac{2GM}{r}}

Given that earth's mass was half its actual value but its radius stayed the same. The escape velocity will be calculated as below:-

V_e=\sqrt{\dfrac{2GM}{r\times 2}}

V_e = \sqrt{\dfrac{GM}{r}}.

Therefore, If the earth's mass were half its actual value but its radius stayed the same, the escape velocity of the earth would be V_e = \sqrt{\dfrac{GM}{r}}.

To know more about escape velocity follow

brainly.com/question/14042253

#SPJ4

8 0
1 year ago
Two buses are driving along parallel freeways that are 5mi apart, one heading east and the other heading west. Assuming that eac
Oksanka [162]

Answer:

101.54m/h

Explanation:

Given that the buses are 5mi apart, and that they are both driving at the same speed of 55m/h, rate of change of distance can be determined using differentiation as;

Let l be the be the distance further away at which they will meet from the current points;

l=\sqrt{13^2-5^2}=12m\\\\\frac{dl}{dt}=-(55m/h+55m/h})\\\\=-110m/h#The speed toward each other.

\frac{dh}{dt}=0, \ \ \ \ h=constant\\\\h^2+l^2=b^2\\\\2h\frac{dh}{dt}+2l\frac{dl}{dt}=2b\frac{db}{dt}\\\\2\times5\times0+2\times12\times(-110)=2\times13\frac{db}{dt}\\\\\frac{db}{dt}=-101.54m/h

Hence, the rate at which the distance between the buses is changing when they are 13mi apart is 101.54m/h

4 0
3 years ago
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