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muminat
3 years ago
8

A neutron star consists of neutrons at approximately nuclear density. Estimate, for a 10-km-diameter neutron star its mass numbe

r, its mass, the acceleration of gravity at its surface.
Physics
1 answer:
slava [35]3 years ago
4 0

Answer:

a) 7.2*10^55

b) 1.2*10^29 kg

c) 3.2*10^11 m/s

Explanation:

Given a diameter of 10 km, that's a radius of 5 km.

We first attempt to find its mass number

A = (r/1.2*10^-15)³

A = (5/1.5*10^-15)³

A = 7.2*10^55

This gotten mass number is then multiplied by the mass of each neutron in the star. With each neutron being, 1.7*10^-27

M = 7.5*10^55 * 1.7*10^-27

M = 1.2*10^29 kg

And finally, using the formula

g = GM/r², we can find its acceleration due to gravity.

g = (6.67*10^-11 * 1.2*10^29) / 5000²

g = 8*10^18 / 2.5*10^7

g = 3.2*10^11 m/s

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5 0
3 years ago
If a ping pong ball and a golf ball are both moving in the same direction with the same amount of kinetic energy, the speed of t
Liono4ka [1.6K]

If the kinetic energy of each ball is equal to that of the other,
then

(1/2) (mass of ppb) (speed of ppb)² = (1/2) (mass of gb) (speed of gb)²

Multiply each side by 2:

      (mass of ppb) (speed of ppb)² = (mass of gb) (speed of gb)²

Divide each side by (mass of gb) and by (speed of ppb)² :

     (mass of ppb)/(mass of gb)  =  (speed of gb)²/(speed of ppb)²

Take square root of each side:

       √ (ratio of their masses)  =  ( 1 / ratio of their speeds)²

By trying to do this perfectly rigorously and elegantly, I'm also
using up a lot of space and guaranteeing that nobody will be
able to follow what I have written.  Let's just come in from the
cold, and say it the clear, easy way:

If their kinetic energies are equal, then the product of each
mass and its speed² must be the same number.

If one ball has less mass than the other one, then the speed²
of the lighter one must be greater than the speed² of the heavier
one, in order to keep the products equal.

The pingpong ball is moving faster than the golf ball.

The directions of their motions are irrelevant.

5 0
3 years ago
A 2,000 kg rocket is launched 12 km straight up at a constant acceleration into the sky at which point the rocket is travelling
hodyreva [135]

Answer:

797700000 J

Explanation:

From the question,

The work done by the rocket, is given as,

W = Ek+Ep............. Equation 1

Where Ek and Ep are the potential and the kinetic energy of the rocket respectively.

Ep = mgh............ Equation 2

Ek = 1/2mv²............. equation 3

Substitute equation 2 and equation 3 into equation 1

W = mgh+1/2mv².............. Equation 4

Where m = mass of the rocket, h = height, v = velocity of the rocket, g = acceleration due to gravity.

Given: m = 2000 kg, h = 12 km = 12000 m, v = 750 m/s, g = 9.8 m/s²

Substitute into equation 4

W = 2000(12000)(9.8)+1/2(2000)(750²)

W = 235200000+562500000

W = 797700000 J

4 0
3 years ago
An alien spaceship traveling at 0.600 c toward the Earth launches a landing craft. The landing craft travels in the same directi
Arturiano [62]

The kinetic energy as measured in the Earth reference frame is 6.704*10^22 Joules.

To find the answer, we have to know about the Lorentz transformation.

<h3>What is its kinetic energy as measured in the Earth reference frame?</h3>

It is given that, an alien spaceship traveling at 0.600 c toward the Earth, in the same direction the landing craft travels with a speed of 0.800 c relative to the mother ship. We have to find the kinetic energy as measured in the Earth reference frame, if the landing craft has a mass of 4.00 × 10⁵ kg.

                  V_x'=0.8c\\V=0.6c\\m=4*10^5kg

  • Let us consider the earth as S frame and space craft as S' frame, then the expression for KE will be,

                  KE=m_0c^2=\frac{mc^2}{1-(\frac{v_x^2}{c^2} )}

  • So, to V_x=(0.8+0.6)c-[\frac{0.6c*(0.8c)^2}{c^2}]=1.016find the KE, we have to find the value of speed of the approaching landing craft with respect to the earth frame.
  • We have an expression from Lorents transformation for relativistic law of addition of velocities as,

                      V_x'=\frac{V_x-V}{1-\frac{VV_x}{c^2} } \\thus,\\V_x=V_x'(1-\frac{VV_x}{c^2} )+V

  • Substituting values, we get,

          V_x=0.8c(1-\frac{0.8c*0.6c}{c^2} )+0.6c=(0.8c*0.52)+0.6c=1.016c

  • Thus, the KE will be,

              KE=\frac{4*10^5*(3*10^8)^2}{\sqrt{1-\frac{(1.016c)^2}{c^2} } } =\frac{1.2*10^{22}}{0.179}=6.704*10^{22}J

Thus, we can conclude that, the kinetic energy as measured in the Earth reference frame is 6.704*10^22 Joules.

Learn more about frame of reference here:

brainly.com/question/20897534

SPJ4

3 0
2 years ago
The total charge a battery can supply is rated in mA⋅h , the product of the current (in mA ) and the time (in h ) that the batte
natita [175]

Answer: 0.2  hours

Explanation: In order to solve this question we have to considerer that a recargeable battery can supply 1800 mA  in one hour then we have to determine how long could this battery drive current through a long, thin wire of resistance 34 Ω .

Besides, this battery has a voltage of 12 V

so by using the Ohm law we also know that V=R*I,

Fron this we can obtain:

I= V/R= 12 V/ 34 Ω=0.35 A= 350 mA

then considering that this battery can supply 1800 mA in one hour we have this battery can supply 350 mA  in x time in the form:

1hour------- 1800 mA

x hour--------350 mA

time= 350/1800= 0.2 hour

4 0
3 years ago
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