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Irina-Kira [14]
2 years ago
8

A student in your class whom you kind-of like asks you come to watch a meteor shower. What exactly are you being invited to? a.

something your mom would not approve of, involving streams of dirty water b. the closest approach of an active bright comet to the Earth c. a once-in-a-lifetime experience at the end of which you will be completely covered with dust particles from space d. watching the left-over dirt from a comet burn up by friction as the pieces hit the Earth's atmosphere e. something quite dangerous where you are likely to be hit when many larger rocks fall from above Submit Your Answer
Physics
1 answer:
madam [21]2 years ago
5 0

Answer:D. Watching the left-over dirt from a comet burn up by friction as the pieces hit the Earth's atmosphere

Explanation: A meteor shower is giant rocks moving across the sky

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Answer:

F = 0.00156[N]

Explanation:

We can solve this problem by using Newton's proposed universal gravitation law.

F=G*\frac{m_{1} *m_{2} }{r^{2} } \\

Where:

F = gravitational force between the moon and Ellen; units [Newtos] or [N]

G = universal gravitational constant = 6.67 * 10^-11 [N^2*m^2/(kg^2)]

m1= Ellen's mass [kg]

m2= Moon's mass [kg]

r = distance from the moon to the earth [meters] or [m].

Data:

G = 6.67 * 10^-11 [N^2*m^2/(kg^2)]

m1 = 47 [kg]

m2 = 7.35 * 10^22 [kg]

r = 3.84 * 10^8 [m]

F=6.67*10^{-11} * \frac{47*7.35*10^{22} }{(3.84*10^8)^{2} }\\ F= 0.00156 [N]

This force is very small compare with the force exerted by the earth to Ellen's body. That is the reason that her body does not float away.

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2 years ago
A ball is dropped from rest from a height h above the ground. another ball is thrown vertically upwards from the ground at the i
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The position of the first ball is

y_1=h-\dfrac g2t^2

while the position of the second ball, thrown with initial velocity v, is

y_2=vt-\dfrac g2t^2

The time it takes for the first ball to reach the halfway point satisfies

\dfrac h2=h-\dfrac g2t^2

\implies\dfrac h2=\dfrac g2t^2

\implies t=\sqrt{\dfrac hg}

We want the second ball to reach the same height at the same time, so that

\dfrac h2=v\sqrt{\dfrac hg}-\dfrac g2\left(\sqrt{\dfrac hg}\right)^2

\implies h=2v\sqrt{\dfrac hg}-g\left(\dfrac hg\right)

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The answer is letter a

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:) What is practical machine? what is the reda<br>tion between MA and VR in a practical<br>machine?​
denis-greek [22]
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a fly wheel of mass 12kg and radius 0.5m with wrapped around it.on the end of the rope is 2kg load.intially it is at rest.calcul
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