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3 years ago

Find the radius to which the sun must be compressed for it to become a black hole.

Physics

1 answer:

ankoles [38]3 years ago

7 0

Answer:

2950 m

Explanation:

The radius to which the sun must be compressed to become a black hole is equal to its Schwarzschild radius, which is given by the formula:

$r=\frac{2GM}{c^2}$

where

G is the gravitational constant

M is the mass of the star

c is the speed of light

For the Sun, its mass is

$M=1.99\cdot 10^{30}kg$

Therefore, its Schwarzschild radius is

$r=\frac{2(6.67\cdot 10^{-11}Nm^2kg^{-2}(1.99\cdot 10^{30} kg)}{(3\cdot 10^8 m/s)^2}=2950 m$

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Anything that has mass and occupies space is called

Jlenok [28]

Anything that has mass and occupies space is called matter.
Matter is of three types:

Solid
Liquid
Gas

Answer:

Matter

Hope you could get an idea from here.

Doubt clarification - use comment section.

8 0

3 years ago

A cycle travels along a circular track of diameter 42 m. Calculate the distance travelled and the displacement of the cycle in (

DENIUS [597]

Answer:

(a) i) The distance travelled by the cycle in half round is approximately 65.97 m

ii) The displacement is 42 m

(b) (i) The distance travelled in one round is approximately 131.95 m

(ii) The displacement of the cycle in one round is 0

Explanation:

The diameter of the track through which the cycle travels, D = 42 m

(a) i) Half round is the motion of half the length of the circular path

The distance travelled by the cycle in half round = The length of half the circular track = (1/2) × π × D

∴ The distance travelled by the cycle in half round = (1/2) × π × 42 m = 21·π m ≈ 65.97 m

ii) The displacement half round = The change in the location of the cycle = The difference between the start and stop locations of the cycle on a straight line after half round

The angle at the center of the circular path the cycle turns in half round = 180°

Therefore, the path between the start and stop location of the cycle in half round = The diameter of the circular track

The displacement of the cycle in half round = The diameter of the circular track = 42 m

(b) (i) The distance travelled in one round = The perimeter of the circular track = π·D

∴ The distance travelled in one round = π × 42 m ≈ 131.95 m

(ii) The displacement of the cycle in one round = The change in the location of the cycle

The start and stop location of the cycle after moving one round is the same, therefore, there is no change in the location of the cycle.

Therefore we have;

The displacement of the cycle in one round = 0 (no change in location of the cycle)

7 0

3 years ago

Is energy transformation from potential to kinetic 100%?

Arada [10]

Yes. take a bow for instance. while pulling back the string you have potential energy. when you let the string go and the arrow flies towards your target the string is filled with kinetic energy.

8 0

3 years ago

A wave of amplitude 20mm has intensity Ix. Another wave of the same frequency but of amplitude 5mm has an intensity Iy.

Alexeev081 [22]

Answer:

(C) 16

Explanation:

Given:

The amplitude of first wave (s₁) = 20 mm

The amplitude of second wave (s₂) = 5 mm

Intensity of first wave = Iₓ

Intensity of second wave = $I_y$

The intensity associated with a wave depends on the amplitude of the wave.

The intensity (I) is directly proportional to the square of the amplitude (s) of the wave and is expressed as:

$I=ks^2\\Where\ k\to constant\ of\ proportionality$

Now, the intensities of the two waves are given as:

$I_x=ks_1^2=k(20)^2\\\\I_y=ks_2^2=k(5)^2$

Dividing both the intensities, we get:

$\frac{I_x}{I_y}=\frac{k(20)^2}{k(5)^2}\\\\\frac{I_x}{I_y}=\frac{400}{25}\\\\\frac{I_x}{I_y}=16$

Therefore, the option (C) is correct.

5 0

3 years ago

Determine the frequency of the 2nd harmonic of a spring that has a 3rd harmonic resonance of f3=512 Hz.

Elena L [17]

Answer:

1) 341 Hz

Explanation:

When a string vibrates, it can vibrate with different frequencies, corresponding to different modes of oscillations.

The fundamental frequency is the lowest possible frequency at which the string can vibrate: this occurs when the string oscillate in one segment only.

If the string oscillates in n segments, we say that it is the n-th mode of vibration, or n-th harmonic.

The frequency of the n-th harmonic is given by

$f_n = nf_1$