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3 years ago

Beyond what point must an object be squeezed for it to become a black hole

Physics

1 answer:

Naddik [55]3 years ago

4 0

Answer:

Condition

$\frac{m}{R} \geq \frac{c^{2} }{2*g}$

Explanation:

Basically black hole is an object from which light rays can not escape it means to go out from gravitational field , that body should thrown with speed greater then light.

Let's do some calculation

Gravitational potential at surface = $-\frac{G*M*m}{R}$

If we give kinetic energy equal to magnitude of Potential energy as on surface it will escape.

$\frac{m*v^{2} }{2}$ = $-\frac{G*M*m}{R}$

⇒ $\frac{m}{R} =\frac{c^{2} }{2*G}$

It will be more better for black hole if above ratio (analogous to density ) is more then above calculated

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A river has a steady speed of 0.550 m/s. A student swims upstream a distance of 1.00 km and swims back to the starting point. If

nlexa [21]

Answer:

$t=564.83seconds=9.414minutes=0.1569hours$

Explanation:

Here the steady speed of the river water relative to ground is

$V_{wg}=0.55m/s$

Speed of the boy relative to still water is

$V_{BW}=1.50m/s$

Here we are considering +ve speed along along +ve x-axis and -ve speed along -ve x-axis

Therefore the speed of boy relative to the ground upstream is:

$V_{BGup}=V_{BW}-V_{WG}\\V_{BGup}=1.50m/s-0.550m/s\\V_{BGup}=0.95m/s$

And the speed of boy relative to the ground downstream is:

$V_{BGdown}=V_{BW}+V_{WG}\\V_{BGdown}=1.50m/s+0.550m/s\\V_{BGdown}=2.05m/s$

The distance covered in upstream trip d₁=1.0km=1000m and the distance covered in downstream is d₂=1.0km=1000m

Since time taken by a person is to cover distance d with speed v given by:

$t=d/v$

The total time taken for one trip is:

$t=\frac{d_{1} }{V_{BGup} } -\frac{d_{2} }{V_{BGdown} }\\t=\frac{1000m}{0.95m/s}-\frac{1000m}{2.05m/s}\\ t=564.83seconds=9.414minutes=0.1569hours\\$

3 0

3 years ago

What happens to the particles of matter vibrated by a wave after the wave passes through the matter

Mariulka [41]

Answer:

As they vibrate, they pass the energy of the disturbance to the particles next to them, which pass the energy to the particles next to them, and so on.

Explanation:

5 0

3 years ago

3. A hydraulic lift is used to jack a 950-kg car 45 cm off the floor. The diameter of the output piston is 23 cm, and the input

ivolga24 [154]

Answer:

a. A input = 0.001669 m²

b. W= 4.193775 KJ

c. h output = 0.60357 cm

d. n = 74.55638 strokes

e. 4.1937 N = 4.1937 N

Explanation:

Hydraulic jacks lift loads using the force created by the pressure in the cylinder chamber by applying small effort. It works on Pascal's principles which explains that the pressure at a certain level and through a mass of fluid at rest is the same in all the directions.

Parameters given:

Diameter of the output piston, d = 0.23 m

Mass of the car, m= 950 kg

Force applied at the input piston, f = 375 N

Height, h = 0.45 m

(a) Finding the area of the input piston:

First, we use Pascal's principle to find the area

(f ÷ A output) ÷ (f ÷ A input)

Where A= area

g = 9.81

A output = πd² ÷ 4

(f ÷ (πd² ÷ 4)) = (f ÷ A input)

[(950 x 9.81) ÷ ((3.14 x 0.23²) ÷ 4) ] = 375 ÷ A input

9319.5 ÷ 0.0415 = 375 ÷ A input

A input = 0.001669 m²

(b) Finding the work done in lifting the car 45 cm

Work done, W = force, f x distance ( which in this case is height, h)

= (950 x 9.81) x 0.45

W= 4.193775 KJ

(c) Finding how the car move up for each stroke if the input piston moves 15 cm in each stroke.

W output = W input

F x h output = F x h input

= (950 x 9.81) x h output = 375 x 0.15

h output = 0.0060357 m

h output = 0.60357 cm

(d) Finding the number of strokes that are required to jack the car up 45 cm

n = h ÷ h output

n = 45 ÷ 0.60357 cm

n = 74.55638 strokes

(e) How the energy is conserved

W output = W input

F x h output = F x h input x n

(950 x 9.81) x 0.45 = 375 x 0.15 x 74.55638

4.1937 N = 4.1937 N

6 0

3 years ago

A 2000 kg test car travels around a circular track with a radius of 100 m. If the

Anarel [89]

Answer:

V=0.38 m/s

Explanation:

ac=3 m/s^2

m=2000kg

r=100m

ac=mv^2/r

Solve for v^2

3=2000(v^2)/100

3=20v^2

3/20=.15

sqrt of .15= 0.38 m/s

V=0.38 m/s

6 0

3 years ago

What is the sequence of energy transformations from the moment the ball is dropped to the moment the board is bent to its maximu

Mila [183]

Answer:

Explanation:

Transformation of energy involves conversion of energy from one form to another for example our movement around involves the conversion of chemical energy stored in the food we eat to other forms of energy such as kinetic energy for the movement, electrical energy in the neurons for impulses and others

The ball posses gravitational potential energy since it is held at a displacement to the ground ( zero point) and when released, the gravitational potential energy is converted to kinetic energy which leads to the fall of the ball until it is at zero displacement to the earth. The board likewise when bent to its maximum extent stored elastic potential energy as a result of the partial displacement of its constituent particle provided it is not stretch beyond its elastic limit which can lead to deformation of the board and the elastic potential energy lost.

5 0

4 years ago