Net force = ?

What's at the bottom of the black hole? Explain .

Brilliant_brown [7]

Answer:

Unknown

Explanation:

By definition, we can't observe what's inside there, because no light – no information of any kind – can escape a black hole. But astrophysical theories suggest that, at the core of a black hole, all the black hole's mass is concentrated into a tiny point of infinite density. This point is known as a singularity.

7 0

3 years ago

An engine absorbs 1.69 kJ from a hot reservoir at 277°C and expels 1.25 kJ to a cold reservoir at 27°C in each cycle.

Anna35 [415]

Answer

Given,

Energy absorbed, $Q_H = 1.69\ kJ$

Energy expels, $Q_C = 1.25\ kJ$

Temperature of cold reservoir, T = 27°C

a) Efficiency of engine

$\eta = \dfrac{Q_H - Q_C}{Q_H}\times 100$

$\eta = \dfrac{1.69 - 1.25}{1.69}\times 100$

$\eta =26.03 %$

b) Work done by the engine

$W = Q_H- Q_C$

$W =1.69 - 1.25$

$W = 0.44\ kJ$

c) Power output

t = 0.296 s

$P = \dfrac{W}{t}$

$P = \dfrac{0.44}{0.296}$

$P = 1.486\ kW$

8 0

3 years ago

A beam of sunlight falling on a prism refracts and forms seven color bands. This illustrates that

ExtremeBDS [4]

That isn"t the right answer the correct answer is B.

4 0

3 years ago

Read 2 more answers

A U-shaped tube open to the air at both ends contains some mercury. A quantity of water is carefully poured into the left arm of

Gekata [30.6K]

Answer:

a) $P=2450\ Pa$

b) $\delta h=23.162\ cm$

Explanation:

Given:

height of water in one arm of the u-tube, $h_w=25\ cm=0.25\ m$

a)

Gauge pressure at the water-mercury interface,:

$P=\rho_w.g.h_w$

we've the density of the water $=1000\ kg.m^{-3}$

$P=1000\times 9.8\times 0.25$

$P=2450\ Pa$

b)

Now the same pressure is balanced by the mercury column in the other arm of the tube:

$\rho_w.g.h_w=\rho_m.g.h_m$

$1000\times 9.8\times 0.25=13600\times 9.8\times h_m$

$h_m=0.01838\ m=1.838\ cm$

<u>Now the difference in the column is :</u>

$\delta h=h_w-h_m$

$\delta h=25-1.838$

$\delta h=23.162\ cm$

7 0

3 years ago

A large power plant heats 1917 kg of water per second to high-temperature steam to run its electrical generators.

erastova [34]

Complete Question

A large power plant heats 1917 kg of water per second to high-temperature steam to run its electrical generators.

(a) How much heat transfer is needed each second to raise the water temperature from 35.0°C to 100°C, boil it, and then raise the resulting steam from 100°C to 450°C? Specific heat of water is 4184 J/(kg · °C), the latent heat of vaporization of water is 2256 kJ/kg, and the specific heat of steam is 1520 J/(kg · °C).

J

(b) How much power is needed in megawatts? (Note: In real power plants, this process occurs under high pressure, which alters the boiling point. The results of this problem are only approximate.)

MW

Answer:

The heat transferred is $Q = 5.866 * 10^9 J$