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2 years ago

What is a Super Massive Black Hole?

Physics

1 answer:

Serggg [28]2 years ago

4 0

Answer:

I dont really know much but i know that it swallow anything it comes across in space.

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When an object moved in a circle ___ acts to accelerate the object toward the center of that cirlce

Irina18 [472]

Centripital acceleration. When an object moves in a circle, centripital acceleration acts to accelerate the object towards the center of that circle

7 0

2 years ago

Read 2 more answers

A soccer player kicks a ball horizontally at 28 m/s from a bridge and hits the ground after 1.5 s. What is the range of the ball

Lelu [443]

Answer:

Explanation:

range of ball = u×t

= 28×1.5

=42

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2 years ago

A rod of length Lo moves iwth a speed v along the horizontal direction. The rod makes an angle of (θ)0 with respect to the x' ax

Colt1911 [192]

Answer:

From the question we are told that

The length of the rod is $L_o$

The speed is v

The angle made by the rod is $\theta$

Generally the x-component of the rod's length is

$L_x = L_o cos (\theta )$

Generally the length of the rod along the x-axis as seen by the observer, is mathematically defined by the theory of relativity as

$L_xo = L_x \sqrt{1 - \frac{v^2}{c^2} }$

=> $L_xo = [L_o cos (\theta )] \sqrt{1 - \frac{v^2}{c^2} }$

Generally the y-component of the rods length is mathematically represented as

$L_y = L_o sin (\theta)$

Generally the length of the rod along the y-axis as seen by the observer, is also equivalent to the actual length of the rod along the y-axis i.e $L_y$

Generally the resultant length of the rod as seen by the observer is mathematically represented as

$L_r = \sqrt{ L_{xo} ^2 + L_y^2}$

=> $L_r = \sqrt{[ (L_o cos(\theta) [\sqrt{1 - \frac{v^2}{c^2} }\ \ ]^2+ L_o sin(\theta )^2)}$

=> $L_r= \sqrt{ (L_o cos(\theta)^2 * [ \sqrt{1 - \frac{v^2}{c^2} } ]^2 + (L_o sin(\theta))^2}$

=> $L_r = \sqrt{(L_o cos(\theta) ^2 [1 - \frac{v^2}{c^2} ] +(L_o sin(\theta))^2}$

=> $L_r = \sqrt{L_o^2 * cos^2(\theta) [1 - \frac{v^2 }{c^2} ]+ L_o^2 * sin(\theta)^2}$

=> $L_r = \sqrt{ [cos^2\theta +sin^2\theta ]- \frac{v^2 }{c^2}cos^2 \theta }$

=> $L_o \sqrt{1 - \frac{v^2}{c^2 } cos^2(\theta ) }$

Hence the length of the rod as measured by a stationary observer is

$L_r = L_o \sqrt{1 - \frac{v^2}{c^2 } cos^2(\theta ) }$

Generally the angle made is mathematically represented

$tan(\theta) = \frac{L_y}{L_x}$

=> $tan {\theta } = \frac{L_o sin(\theta )}{ (L_o cos(\theta ))\sqrt{ 1 -\frac{v^2}{c^2} } }$

=> $tan(\theta ) = \frac{tan\theta}{\sqrt{1 - \frac{v^2}{c^2} } }$

Explanation:

7 0

3 years ago

8 th Grade Physical Science Worksheet Part A: Read the scenario and answer the questions. David read that Fox brake pads and Bes

svet-max [94.6K]

Okokokokokokokokok!OK?

4 0

2 years ago

How does increasing the amount of charge on an object affect the electric force it exerts on another charged object?

gregori [183]

Hi Mandy!

Question - How does increasing the amount of charge on an object affect the electric force it exerts on another charged object?

Answer - The electric force increases because the amount of charge has a direct relationship to the force.

Why - The affect that the force exerts on another object that is also charge is becase the fact that when the electric force increases is when the charge is direct with the object with the force.

Hope This Helps :)

4 0

3 years ago