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3 years ago

Can someone explain which of Newton’s Law is demonstrated in part 1 and which is demonstrated in part 2? (Picture)

Physics

1 answer:

rewona [7]3 years ago

3 0

Answer:

Every action has an equal and opposite reaction. If the student doesn't push, nothing moves, is one student pushes, both move which is an example of newtons third law.

Explanation:

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A major source of water pollution comes from that washes chemicals and other pollutants from improperty managed land.

IrinaK [193]

The answer to this would be B

4 0

3 years ago

When two point charges are a distance d part, the electric force that each one feels from the other has magnitude F. In order to

Serjik [45]

Answer:

E) d/sqrt2

Explanation:

The initial electric force between the two charge is given by:

$F=k\frac{q_1 q_2}{d^2}$

where

k is the Coulomb's constant

q1, q2 are the two charges

d is the separation between the two charges

We can also rewrite it as

$d=\sqrt{k\frac{q_1 q_2}{F}}$

So if we want to make the force F twice as strong,

F' = 2F

the new distance between the charges would be

$d'=\sqrt{k\frac{q_1 q_2}{(2F)}}=\frac{1}{\sqrt{2}}\sqrt{k\frac{q_1 q_2}{(2F)}}=\frac{d}{\sqrt{2}}$

so the correct option is E.

8 0

3 years ago

PLEASE PLEASE HELPPPPPP!!!!!!!’

Ostrovityanka [42]

The answer is B

explanation

3 0

3 years ago

Saturn has an orbital period of 29.46 years. In two or more complete sentences, explain how to calculate the average distance fr

vivado [14]

This question can be solved from the Kepler's law of planetary motion.

As per this law the square of time period of a planet is proportional to the cube of semi major axis.

Mathematically it can be written as $T^{2} \alpha R^{3}$

⇒ $T^{2} = KR^{3}$

Here K is the proportionality constant.

If $T_{1}$ and $T_{2}$ are the orbital periods of the planets and

$R_{1}$ and $R_{2}$ are the distance of the planets from the sun, then Kepler's law can be written as-

$\frac{T_{1} ^{2} }{T_{2} ^{2} } =\frac{R_{1} ^{3} }{R_{2} ^{2} }$

⇒ $R_{1} ^{3} =R_{2} ^{3} *\frac{T_{1} ^{2} }{T_{2} ^{2} }$

Here we are asked to calculate the the distance of Saturn from sun.It can solved by comparing it with earth.

Let the distance from sun and orbital period of Saturn is denoted as $R_{1}$ and $T_{1}$ respectively.

Let the distance from sun and orbital period of earth is denoted as $R_{2}$ and $T_{2}$ respectively.

we are given that $T_{1} =29.46 years$

we know that $R_{2} =$ 1 AU and $T_{2} =$ 1 year.

1 AU is the mean distance of earth from the sun which is equal to 150 million kilometre.

Hence distance of Saturn from sun is calculated as -

From Kepler's law as mentioned above-

$R_{1} ^{3} =R_{2} ^{3} *\frac{T_{1} ^{2} }{T_{2} ^{2} }$

= $[1 ]^{3} *\frac{[29.46]^{2} }{[1]^{2} } AU$

$=867.8916 AU^{3}$

⇒ $R_{1} =\sqrt[3]{867.8916}$

=9.5386 AU [ans]

5 0

3 years ago

A is a process that involves reactants and products

dusya [7]

I think the answer is chemical reactions.

6 0

4 years ago