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2 years ago

Choose a system consisting only of ball A. What is the momentum change of the system during the collision?

Physics

1 answer:

shusha [124]2 years ago

3 0

The momentum change of the system during the collision is always equals to one another.

<h3>Momentum change during collision</h3>

In a collision, the momentum change of first object is equal to and opposite of the momentum change of second object with which the first one collided. Or in other words, the momentum lost by first object is equal to the momentum gained by second object.

The total momentum of the system means the collection of two objects is conserved so we can conclude that the momentum change of the system during the collision is always equals and opposite to one another.

Learn more about collisions here: brainly.com/question/7538238

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Which can be observed both on Earth and in an accelerating ship in space that is free from the effect of any gravitational field

Reil [10]

Answer:

apples falling from trees
people's feet touching the ground
sky divers moving toward the ground
balls bending downward after being thrown

Explanation:

When a space ship is accelerating in space, there is a force known as Inertia that kicks in. Inertia will mirror the effects of gravity on the ship even if there is no gravitational field effect such that anything that would happen where there is gravity, would continue to happen.

This means that apples will fall from trees, people's feet will touch the ground, sky divers will be pulled downwards and balls will bend downwards when thrown as well. These are the same effects expected on earth where gravity pulls things towards the earth's core.

8 0

2 years ago

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The electric field must be zero inside a conductor in electrostatic equilibrium, but not inside an insulator. It turns out that

pav-90 [236]

Answer:

The permittivity of rubber is $\epsilon = 8.703 *10^{-11}$

Explanation:

From the question we are told that

The magnitude of the point charge is $q_1 = 70 \ nC = 70 *10^{-9} \ C$

The diameter of the rubber shell is $d = 32 \ cm = 0.32 \ m$

The Electric field inside the rubber shell is $E = 2500 \ N/ C$

The radius of the rubber is mathematically evaluated as

$r = \frac{d}{2} = \frac{0.32}{2} = 0.16 \ m$

Generally the electric field for a point is in an insulator(rubber) is mathematically represented as

$E = \frac{Q}{ \epsilon } * \frac{1}{4 * \pi r^2}$

Where $\epsilon$ is the permittivity of rubber

=> $E * \epsilon * 4 * \pi * r^2 = Q$

=> $\epsilon = \frac{Q}{E * 4 * \pi * r^2}$

substituting values

$\epsilon = \frac{70 *10^{-9}}{2500 * 4 * 3.142 * (0.16)^2}$

$\epsilon = 8.703 *10^{-11}$

7 0

3 years ago

Electrons are allowed "in between" quantized energy levels, and, thus, only specific lines are observed. The energies of atoms a

lawyer [7]

Answer:

This is because The energies of atoms are quantized.

Electrons are allowed "in between" quantized energy levels, and, thus, only specific lines are observed

5 0

3 years ago

What is the difference between speed and velocity

tiny-mole [99]

Speed is the distance travelled by an object whereas velocity is distance travelled by an object per unit time in a given direction.

8 0

3 years ago

The Carson family's pancake recipe uses 2 teaspoons of baking powder for every 1/3 of a teaspoon of salt. How much baking powder

melomori [17]

Answer:

6 teaspoons of baking powder required.

Explanation:

Given that

According to the recipe of pancake,

For every $\frac{1}{3}$ teaspoon of salt, 2 teaspoons of baking baking powder is required.

To find:

How much baking powder will be needed, if 1 teaspoon of salt was used ?

Solution:

This problem can be solved using ratio.

$\frac{1}3$ teaspoon of salt : 2 teaspoons of baking powder

Let us multiply the above ratio with 3.

$\frac{1}{3}\times 3$ teaspoon of salt : 2 $\times$ 3 teaspoons of baking powder

1 teaspoon of salt : 6 teaspoons of baking powder

So, answer is <em>6 teaspoons </em>of baking powder required.

Also, we can use the unitary method:

$\frac{1}3$ teaspoon of salt needs = 2 teaspoons of baking powder

1 teaspoon of salt needs = $\frac{2}{\frac{1}3}$ teaspoons of baking powder

1 teaspoon of salt needs = 2 $\times$ 3 = <em>6</em> teaspoons of baking powder needed

So, the answer is:

<em>6 teaspoons of baking powder </em>required.

5 0

3 years ago

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