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slavikrds [6]
4 years ago
15

A ____ is a region so dense that nothing, including light, can escape its gravity field.

Physics
2 answers:
vazorg [7]4 years ago
8 0

Answer:

black hole

Explanation:

A black hole is a region which is highly dense and very high gravitational field.

As the density of black hole is very high, so the mass of black g=hole is very large thus the force of gravitation is very large. So, even light cannot escape from the gravitational filed from the black hole.

Oksana_A [137]4 years ago
5 0
The answer is c. black hole
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Find the period of a wave with a frequency 40 kHz​
Burka [1]

We know that frequency is an inverse value of time f=\dfrac{1}{t} this implies that time is the inverse value of frequency t=\dfrac{1}{f}.

Now since <em>f</em> is <em>40kHz</em><em>,</em><em> </em>we can calculate period or duration for that matter time.

t=\dfrac{1}{40\mathbf{kHz}}=0.000025\mathbf{s}=25\mathbf{\mu s}

Hope this helps.

r3t40

4 0
3 years ago
Can someone help me please??<br> This question is confusing to me:(
puteri [66]
Its d. the water isn't moving the fish, and that's how they remain stationary. they don't have any currents pushing on them. if this is wrong, its c, which is practically the same thing. but its most likely d.
7 0
4 years ago
Water flowing through a cylindrical pipe suddenly comes to a section of pipe where the diameter decreases to 86% of its previous
Orlov [11]

Answer:

Explanation:

The speed of the water in the large section of the pipe is not stated

so i will assume 36m/s

(if its not the said speed, input the figure of your speed and you get it right)

Continuity equation is applicable for ideal, incompressible liquids

Q the flux of water that is  Av with A the cross section area and v the velocity,

so,

A_1V_1=A_2V_2

A_{1}=\frac{\pi}{4}d_{1}^{2} \\\\ A_{2}=\frac{\pi}{4}d_{2}^{2}

the diameter decreases 86% so

d_2 = 0.86d_1

v_{2}=\frac{\frac{\pi}{4}d_{1}^{2}v_{1}}{\frac{\pi}{4}d_{2}^{2}}\\\\=\frac{\cancel{\frac{\pi}{4}d_{1}^{2}}v_{1}}{\cancel{\frac{\pi}{4}}(0.86\cancel{d_{1}})^{2}}\\\\\approx1.35v_{1} \\\\v_{2}\approx(1.35)(38)\\\\\approx48.6\,\frac{m}{s}

Thus, speed in smaller section is 48.6 m/s

3 0
3 years ago
What is the frequency for a beam of electrons orbiting in a field of 4.62 x 10^-3 T? Let the mass of an electrons m = 9.31 x 10^
melisa1 [442]

Frequency: 1.27\cdot 10^8 Hz

Explanation:

The force experienced by an electron in a magnetic field is

F=qvB

where

q=1.6\cdot 10^{-19}C is the electron charge

v is the speed of the electron

B is the strength of the magnetic field

Since the force is perpendicular to the direction of motion of the electron, the force acts as centripetal force, so we can write:

qvB=m\frac{v^2}{r}

where

r is the radius of the orbit

m=9.31\cdot 10^{-31}kg is the mass of the electron

Re-arranging the equation,

\frac{v}{r}=\frac{qB}{m} (1)

We also know that in a circular motion, the speed is equal to the ratio between circumference of the orbit and orbital period (T):

v=\frac{2\pi r}{T}

Substituting into (1),

\frac{2\pi}{T}=\frac{qB}{m}

We also know that 1/T is equal to the frequency f, so

f=\frac{qB}{2\pi m}

In this problem,

B=4.62\cdot 10^{-3}T

Therefore, the frequency of the electrons is

f=\frac{(1.6\cdot 10^{-19})(4.62\cdot 10^{-3})}{2\pi(9.31\cdot 10^{-31})}=1.27\cdot 10^8 Hz

4 0
3 years ago
A charged box (m=445 g, ????=+2.50 μC) is placed on a frictionless incline plane. Another charged box (????=+75.0 μC) is fixed i
victus00 [196]

The concept required to perform this exercise is given by the coulomb law.

The force expressed according to this law is given by

F= \frac{kqQ}{r^2}

Where,

k = 8.99 * 10^9 N m^2 / C^2.

q = charges of the objects

r= distance/radius

Our values are previously given, so

q= 2.5*10^{-6}C\\Q= 75*10^{-6}C\\r=0.59

Replacing,

F=\frac{kqQ}{r^2}

F= \frac{(8.99 x 10^9)(2.5*10^{-6})(75*10^{-6})}{0.59^2}

F= 4.8423N

The force acting on the block are given by,

F-mgsin\theta = ma

a = \frac{F-mgsin\theta}{m}

a = \frac{4.8423-(0.445)(9.8)sin(35)}{0.445}a = 10.31m/s^2

Therefore the box is accelerated upward.

3 0
3 years ago
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