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postnew [5]
3 years ago
13

1.Explain frame of reference. How is it different when you are riding inside a car or standing by the highway?

Physics
1 answer:
galben [10]3 years ago
8 0

Answer:

See explanation

Explanation:

A frame of reference simply refers to a set of coordinates which is used to determine the position and velocity of objects all the objects found in that particular frame. The person inside the car and standing by the highway are in different frames of reference.

Newtons first law of motion states that an object will continue to be at rest or in a state of uniform velocity until it is acted upon by an external force.

Newtons second law states that rate of change of momentum is directly proportional to the impressed force.

Newtons third law states that action and reaction are equal and opposite.

The action of applying brakes leads to an equal reaction of a sudden forward movement. Seat belts help to prevent the occupants of a car from being injured by the sudden stoppage of the car.

Force is simply defined as a push or pull.

Balanced forces do not lead to acceleration of a body but an unbalanced for leads to acceleration of a body.

strong nuclear force, electromagnetic force, weak nuclear force and gravity

Power is the rate of doing work while week is said to be done when the force applied moves a distance in the direction of the force. Power is defined as work/time.

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7 0
3 years ago
If you were to triple the size of the Earth (R = 3R⊕) and double the mass of the Earth (M = 2M⊕), how much would it change the g
EastWind [94]

Answer:

Decreased by a factor of 4.5

Explanation:

"We have Newton formula for attraction force between 2 objects with mass and a distance between them:

F_G = G\frac{M_1M_2}{R^2}

where G =6.67408 × 10^{-11} m^3/kgs^2 is the gravitational constant on Earth. M_1, M_2 are the masses of the object and Earth itself. and R distance between, or the Earth radius.

So when R is tripled and mass is doubled, we have the following ratio of the new gravity over the old ones:

\frac{F_G}{f_g} = \frac{G\frac{M_1M_2}{R^2}}{G\frac{M_1m_2}{r^2}}

\frac{F_G}{f_g} = \frac{\frac{M_2}{R^2}}{\frac{m_2}{r^2}}

\frac{F_G}{f_g} = \frac{M_2}{R^2}\frac{r^2}{m_2}

\frac{F_G}{f_g} = \frac{M_2}{m_2}(\frac{r}{R})^2

Since M_2 = 2m_2 and r = R/3

\frac{F_G}{f_g} = \frac{2}{3^2} = 2/9 = 1/4.5

So gravity would have been decreased by a factor of 4.5  

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