2 years ago

What do we mean by the event horizon of a black hole?

Physics

1 answer:

Fynjy0 [20]2 years ago

6 0

The correct answer is: C. It is the point beyond which neither light nor anything else can escape.

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A horizontal metal bar oriented east-west drops straight down in a location where the earth's magnetic field is due north. as a

amid [387]

Metal bar is aligned along East- West direction and it is dropped vertically down

So its velocity is along -Z direction,

Now the Earth's magnetic field is towards north so its towards +Y direction

now we have formula for force on a moving charge

$F = q(v X B)$

v = - z direction

B = + y direction

now by the above formula

$F = q(v(-k) X B(j))$

$F = qvB (i)$

So force is towards East

so all positive charge is towards East end

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2 years ago

Eddie and Val observed the picture of an athlete running in a race.

dedylja [7]

Answer:

Your answer would be C <u><em>Hope this helps</em></u>

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2 years ago

Which of the following is false

musickatia [10]

what are the options??

please include all of your question

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2 years ago

Read 2 more answers

What is the greenhouse effect?

Zanzabum

A on that problem :)[email protected]

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A rope of total mass m hnd length L is suspended vertically with an object of mass M suspended from the lower end. Find an expre

pantera1 [17]

Answer:

Part a)

$v = \sqrt{xg + \frac{MLg}{m}}$

Part b)

t = 12 s

Explanation:

Part a)

Tension in the rope at a distance x from the lower end is given as

$T = \frac{m}{L}xg + Mg$

so the speed of the wave at that position is given as

$v = \sqrt{\frac{T}{\mu}}$

here we know that

$\mu = \frac{m}{L}$

now we have

$v = \sqrt{\frac{ \frac{m}{L}xg + Mg}{m/L}$

$v = \sqrt{xg + \frac{MLg}{m}}$

Part b)

time taken by the wave to reach the top is given as

$t = \int \frac{dx}{\sqrt{xg + \frac{MLg}{m}}}$

$t = \frac{1}{g}(2\sqrt{xg + \frac{MLg}{m}})$

$t = \frac{2}{9.8}(\sqrt{(39.2\times 9.8) + \frac{8(39.2)(9.8)}{1}})$

$t = 12 s$

4 0

2 years ago