Which two factors affect the size of the gravitational field?

Quentin is playing baseball and is batting. He swings and hits the ball sending it flying to the outfield. Which is greater- the

Bess [88]

Answer:

The force of the ball on the bat is same as the force of the bat on the ball.

Explanation:

A bat hits the ball and the ball moves to the out filed.

According to the Newton's third law, for every action there is an equal and opposite reaction.

The action and the reaction forces acts on the two different bodies but the magnitude of the force is same.

As the ball is hitted by the bat, the bat exerts the force on the ball and the same force is exerted on the bat by the ball according to the Newton's third law.

So, the force of the ball on the bat is same as the force of the bat on the ball but the direction of force is opposite.

5 0

3 years ago

Astronomy question, <br> what keeps supernova explosions from being seen?

kirill115 [55]

a star that suddenly increases greatly in brightness because of a catastrophic explosion that ejects most of its mass.

4 0

3 years ago

Consider two insulating balls with evenly distributed equal and opposite charges on their surfaces, held with a certain distance

siniylev [52]

Answer:

interest point:

1) Point on the left side

2) Point within the radius r₁ of the first sphere

3) Point between the two spheres

4) point within the radius r₂ of the second sphere

5) Right side point

Explanation:

In this case, the total electric field is the vector sum of the electric fields of each sphere, to simplify the calculation on the line that joins the two spheres

We will call the sphere on the left 1 and it has a positive charge Q with radius r1, the sphere on the right is called 2 with charge -Q with radius r2. The total field is

E_ {total} = E₁ + E₂

E_{ total} = $k \frac{Q}{x_1^2} + k \frac{Q}{x_2^2}$

the bold indicate vectors, where x₁ and x₂ are the distances from the center of each sphere. If the distance that separates the two spheres is d

x₂ = x₁ -d

E total = $k \frac{Q}{x_1^2} - k \frac{Q}{(x_1 - d)^2}$

Let's analyze the field for various points of interest.

1) Point on the left side

in this case

E_ {total} = $k Q \ ( \frac{1}{x_1^2} - \frac{1}{(x_1 +d)2} )$

E_ {total} = $k \frac{Q}{x_1^2}$ $( 1 - \frac{1}{(1 + \frac{d}{x_1} )^2 } )$

We have several interesting possibilities:

* We can see that as the point is further away the field is more similar to the field created by two point charges

* there is a point where the field is zero

E_ {total} = 0

x₁² = (x₁ + d)²

2) Point within the radius r₁ of the first sphere.

In this case, according to Gauus' law, the charge is on the surface of the sphere at the point, there is no charge inside so this sphere has no electric field on its inner point

E_ {total} = $-k \frac{Q}{x_2^2} = -k \frac{Q}{((d-x_1)^2}$