Ask question

Ask question

All categories

3 years ago

11

Under some circumstances, a star can collapse into an extremely dense object made mostly of neutrons and called a neutron star.

The density of a neutron star is roughly 1014 times as great as that of ordinary solid matter. Suppose we represent the star as a uniform, solid, rigid sphere, both before and after the collapse. The star's initial radius was 7.0x10^5 km (comparable to our sun); its final radius is 16 km .If the original star rotated once in 35 days, find the angular speed of the neutron star.

1 answer:

xxTIMURxx [149]3 years ago

4 0

Answer:

The angular speed of the neutron star is $3976.978\ rads^{-1}$

Solution:

As per the question:

Initial radius of the star, r = $7.0\times 10^{5}\ km = 7\times 10^{8}\ m$

Final Radius, r' = 16 km

Period of rotation of the star, T = 35 days = $35\times 24\times 3600 = 3024000\ s$

Now,

To calculate the angular speed of the neutron star:

We use the principle of conservation of angular momentum:

$I\omega = I'\omega'$

where

I = Moment of inertia

$\omega = \frac{2\pi}{T}$

$(7\times 10^{8})^{2}\times \frac{2\pi}{T} = (16000)^{2}\omega'$

$\omega = 3976.978\ rads^{-1}$

You might be interested in

During a free fall Swati was accelerating at -9.8m/s2. After 120 seconds how far did she travel? Use the formula =1/2 *

marta [7]

Distance fallen = 1/2 ( V initial + V final ) *t
We know
a = -9.8 m/s2
t=120s

To find distance fallen, we need to find V final
Use the equation
V final = V initial + a*t
Substitute known values
V final = 0 + (-9.8)(120)
V final = -1176 m/s

Then plug known values to distance fallen equation
Distance fallen = 1/2 ( 0 + 1176 )(120)
= 1/2(1776)(120)
=106,560 m

This way plugging into distance equation is actually the long way. A faster way is to plug the values into
Distance fallen = V initial * t + 1/2(a*t)
We won't need to find V final using another equation.

But anyways, good luck!

4 0

3 years ago

How is an electromagnet different from a bar magnet?

Reptile [31]

The answer for the question is c

5 0

2 years ago

Read 2 more answers

A Red Rider bb gun uses the energy in a compressed spring to provide the kinetic energy for propelling a small pellet of mass 0.

Kay [80]

Answer:

a.6.5025 J

b.6.5025 J

Explanation:

We are given that

Mass of pellet,m=0.27 g= $0.27\times 10^{-3} kg$

1 kg=1000 g

Spring constant,k=1800 N/m

x=8.5 cm= $8.5\times 10^{-2} m$

1m=100 cm

a.Potential energy stored in the compressed spring is given by

P.E= $\frac{1}{2}kx^2$

$P.E=\frac{1}{2}(1800)(8.5\times 10^{-2})^2$

$P.E=6.5025 J$

b.By using law of conservation of energy

P.E of spring=K.E of the pellet

K.E of the pellet=6.5025 J

6 0

3 years ago

On a vacation flight, you look out the window of the jet and wonder about the forces exerted on the window. Suppose the air outs

user100 [1]

Answer:

A) $\Delta P = 14512.5 Pa = 14.512 kPa$

B) F = 1632.65 N

Explanation:

Given details

outside air speed is given as $v_2 = 150 m/s$

since inside air is atmospheric , $v_1 = 0 m/s$

a) By using bernoulli equation between outside and inside of flight

$P_1 + \frac{1}{2} \rho v_1^2 + \rho gh = P_2 + \frac{1}{2} \rho v_2^2 + \rho gh$

$\Delta P = \frac{1}{2} \rho v_2^2 - \frac{1}{2} \rho v_1^2$

$\Delta P = \frac{1}{2} \rho[ v_2^2 -v_1^2]$

$\Delta P = \frac{1}{2} 1.29 [ 150^2 - 0^2]$

$\Delta P = 14512.5 Pa = 14.512 kPa$

b) force exerted on window

Area of window $= 25\times 45 = 1125 cm^2 = 0.1125 m^2$

We know that force is given as

$F = P\times A$

$F = 14512.5 \times 0.1125$

F = 1632.65 N

5 0

3 years ago

Two ice skaters, Paula and Ricardo, initially at rest, push off from each other. Ricardo weighs more than Paula.

sveta [45]

Answer:

the two ice skater have the same momentum but the are in different directions.

Paula will have a greater speed than Ricardo after the push-off.

Explanation:

Given that:

Two ice skaters, Paula and Ricardo, initially at rest, push off from each other. Ricardo weighs more than Paula.

A. Which skater, if either, has the greater momentum after the push-off? Explain.

The law of conservation of can be applied here in order to determine the skater that possess a greater momentum after the push -off

The law of conservation of momentum states that the total momentum of two or more objects acting upon one another will not change, provided there are no external forces acting on them.

So if two objects in motion collide, their total momentum before the collision will be the same as the total momentum after the collision.

Momentum is the product of mass and velocity.

SO, from the information given:

Let represent the mass of Paula with $m_{Pa}$ and its initial velocity with $u_{Pa}$

Let represent the mass of Ricardo with $m_{Ri}$ and its initial velocity with $u_{Ri}$

At rest ;

their velocities will be zero, i.e

$u_{Pa}$ = $u_{Ri}$ = 0

The initial momentum for this process can be represented as :

$m_{Pa}$ $u_{Pa}$ + $m_{Ri}$ $u_{Ri}$ = 0

after push off from each other then their final velocity will be $v_{Pa}$ and $v_{Ri}$

The we can say their final momentum is:

$m_{Pa}$ $v_{Pa}$ + $m_{Ri}$ $v_{Ri}$ = 0

Using the law of conservation of momentum as states earlier.

Initial momentum = final momentum = 0

$m_{Pa}$ $u_{Pa}$ + $m_{Ri}$ $u_{Ri}$ = $m_{Pa}$ $v_{Pa}$ + $m_{Ri}$ $v_{Ri}$

Since the initial velocities are stating at rest then ; u = 0

$m_{Pa}$ (0) + $m_{Pa}$ (0) = $m_{Pa}$ $v_{Pa}$ + $m_{Ri}$ $v_{Ri}$

$m_{Pa}$ $v_{Pa}$ + $m_{Ri}$ $v_{Ri}$ = 0

$m_{Pa}$ $v_{Pa}$ = - $m_{Ri}$ $v_{Ri}$

Hence, we can conclude that the two ice skater have the same momentum but the are in different directions.

B. Which skater, if either, has the greater speed after the push-off? Explain.

Given that Ricardo weighs more than Paula

So $m_{Ri} > m_{Pa}$ ;

Then $\mathsf{\dfrac{{m_{Ri}}}{m_{Pa} }= 1}$

The magnitude of their momentum which is a product of mass and velocity can now be expressed as:

$m_{Pa}$ $v_{Pa}$ = $m_{Ri}$ $v_{Ri}$

The ratio is

$\dfrac{v_{Pa}}{v_{Ri}} =\dfrac{m_{Ri}}{m_{Pa}} = 1$

$v_{Pa} >v_{Ri}$

Therefore, Paula will have a greater speed than Ricardo after the push-off.

6 0

3 years ago