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DaniilM [7]
3 years ago
7

How much would you have to shrink a baseball to a black hole?

Physics
1 answer:
Vesna [10]3 years ago
3 0

The Schwarzchild radius is . . . <em>R = 2GM/c²  .</em>

Just like you, I'm not completely sure what that means.  But it DOES use the mass of a black hole to calculate a radius associated with it, so with 15 Brainly points at stake, it seems like a good-enough formula to use for an answer.

Before I proceed, I really should ask you whether you're talking a softball or a hardball, but again . . . . .

So R = 2GM/c²

G = the  gravitational constant = 6.67 x 10⁻¹¹ N-m²/kg²

M = mass of the baseball = 145 grams = 0.145 kg

c = speed of light = 3 x 10⁸ m/s

R = 2 (6.67 x 10⁻¹¹ m³/kg-s²) (0.145 kg) / (3 x 10⁸ m/s)²

R = (2 x 6.67 x 10⁻¹¹ x 0.145 / 9 x 10¹⁶) (m³-kg-s² / kg-s²-m²)

R = ( 1.9343 x 10⁻¹¹ / 9 x 10¹⁶) (m³-kg-s²/m²-kg-s²)

R = (0.2149 x 10⁻²⁷) meter

<em>R = 2.149 x 10⁻²⁸ meter</em>

For reference: The radius of a Hydrogen atom is 1.2 x 10⁻¹⁰ meter.

So in order to make a black hole out of a baseball, you have to crunch the baseball down to around  0.000000000000000001791 the size of a Hydrogen atom.

Would that be a problem for you ?

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zalisa [80]

After my ythorough researching, the orbital velocity of the Earth compare to the Mercury is relative to each other. In addition, Mercury orbits the Sun and it is within the Earth's orbit as its lesser or inferior. The correct answer to the following given statement above is relative to each other.

 

8 0
3 years ago
A ship is towed through a narrow channel by applying forces to three ropes attached to its bow. Determine the magnitude and orie
Y_Kistochka [10]

This question is solved using an available similar problem as data provided for the forces was not given.

Repeat the same steps outlined for your problem.

Regards.

Answer:

F = 1.598 KN , Q = 90 degree (+ y-axis)

Explanation:

Sum of Forces in x-direction to the left (+)

2 cos (30) + 3cos (60) + F*cos (Q) = F_a   ..... 1

Sum of Forces in y-direction to the up (+)

2 sin (30) + F*sin (Q) - 3 sin (60)  ...... 2

Using Eq 2 and solve:

F*sin (Q) = 1.598 KN

F_min when sin (Q) is max, max possible value of sin(Q) = 1 @ Q = 90 degrees.

Hence,

F_min = 1.598 KN

Using Eq 1 @ Q = 90 degrees and F = 1.598 KN:

F_a = 2 cos (30) + 3cos (60)  = 3.2 KN

6 0
3 years ago
A light source emits light with dominant wavelengths in the range of 650 to 690 nm. what is the principle color of light emitted
dsp73

Answer:

Please see below as the answer is self-explanatory.

Explanation:

  • The visible range extends roughly from 400 nm (violet) to 700 nm (red).
  • Below the violet is the ultra-violet spectrum (with higher energy) and above red, we have the infra-red spectrum.
  • The wavelengths in the range of 650 to 690 nm have red as the dominant color.
6 0
3 years ago
A 8.0\,\text {kg}8.0kg8, point, 0, start text, k, g, end text box is released from rest at a height y_0 =0.25\,\text my 0 ​ =0.2
Ratling [72]

Answer:

μ = 0.125

Explanation:

To solve this problem, which is generally asked for the coefficient of friction, we will use the conservation of energy.

Let's start working on the ramp

starting point. Highest point of the ramp

         Em₀ = U = m h y

final point. Lower part of the ramp, before entering the rough surface

        Em_{f} = K = ½ m v²

as they indicate that there is no friction on the ramp

          Em₀ = Em_{f}

          m g y = ½ m v²

          v = \sqrt{2gy}

we calculate

          v = √(2 9.8 0.25)

           v = 2.21 m / s

in the rough part we use the relationship between work and kinetic energy

          W = ΔK = K_{f} -K₀

as it stops the final kinetic energy is zero

          W = -K₀

The work is done by the friction force, which opposes the movement

          W = - fr x

friction force has the expression

          fr = μ N

let's write Newton's second law for the vertical axis

         N-W = 0

         N = W = m g

we substitute

            -μ m g x = - ½ m v²

           μ = \frac{v^{2} }{2 g x}

Let's calculate

           μ = \frac{2.21^{2}}{2\  9.8\  2.0}

           μ = 0.125

4 0
3 years ago
What is the weight of a<br> 63.7 kg person?
IRINA_888 [86]

Answer:

A person that weighs 63.7 kg weighs 63.7 kg

Explanation:

Too complicated to explain.

6 0
3 years ago
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