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likoan [24]
3 years ago
13

B. how does the separation between galaxies compare to the separation between stars? based on your answer, discuss the likelihoo

d of galactic collisions in comparison to the likelihood of stellar collisions.
Physics
1 answer:
leva [86]3 years ago
4 0
Galaxies are much further apart than stars. This is the reason why they are less likely to collide and the likelihood of galactic collision is much smaller  than the likelihood of stellar collision. Example for galaxy collision is the collision of the Milky Way galaxy with Andromeda. It is estimated that the collision will be <span>in about 4.5 billion years. </span> 
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A 20 kg shopping cart moving at a velocity of 0.5 m/s collides with a store wall and
MAXImum [283]

Answer:

<h2>10 kg.m/s</h2>

Explanation:

The momentum of an object can be found by using the formula

momentum = mass × velocity

From the question we have

momentum = 20 × 0.5

We have the final answer as

<h3>10 kg.m/s</h3>

Hope this helps you

7 0
2 years ago
A boy uses a force to keep a rock from falling. (He is carrying the rock.) Some forces, like this one, act only when two objects
MArishka [77]
Gravity is a non contact force , jump from a little high place (don't do that :P) , you wont just get stuck in the air you will fall down , this is gravity you dont need any contact 
5 0
3 years ago
Matter is made up of tiny particles which can be atoms,molecules, or electrically charged particles called?.
yuradex [85]

Answer: alpha particle. i think

Explanation:

6 0
2 years ago
How does the work required to accelerate a particle from 10 m/s to 20 m/s compare to that required to accelerate it from 20 m/s
poizon [28]

To solve this problem we will apply the energy conservation theorem for which the work applied on a body must be equivalent to the kinetic energy of this (or vice versa) therefore

W = \Delta KE

\Delta W = \frac{1}{2} (m)(v_f)^2 -\frac{1}{2} (m)(v_i)^2

Here,

m = mass

v_{f,i} = Velocity (Final and initial)

First case) When the particle goes from 10m/s to 20m/s

\Delta W = \frac{1}{2} (m)(v_f)^2 -\frac{1}{2} (m)(v_i)^2

\Delta W = \frac{1}{2} (m)(20)^2 -\frac{1}{2} (m)(10)^2

W_1 = 150(m) J

Second case) When the particle goes from 20m/s to 30m/s

\Delta W = \frac{1}{2} (m)(v_f)^2 -\frac{1}{2} (m)(v_i)^2

\Delta W = \frac{1}{2} (m)(30)^2 -\frac{1}{2} (m)(20)^2

W_1 = 250(m) J

As the mass of the particle is the same, we conclude that more energy is required in the second case than in the first, therefore the correct answer is A.

5 0
3 years ago
How does an increase in cold working effect Modulus of Elasticity and why?
VladimirAG [237]

Answer:

There is a decrease in modulus of elasticity

Explanation:

Young's Modulus of elasticity also known as elastic modulus is the deformation of a body along a particular axis under the action of opposing forces along that axis. at atomic levels, it depends on bond energy or strength.

In cold working processes, plastic deformation a metal occurs below its re-crystallization temperature due to which crystal structure of metal gets distorted and as a result of dislocations fractures also occur resulting in hardening of metal but bonds at atomic levels defining elasticity are temporarily affected.

Thus an increase in cold working results in a decrease in modulus of elasticity.

7 0
3 years ago
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