Ask question

Ask question

All categories

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.

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

You might be interested in

You see lightning and 30 seconds later you hear thunder. how far away is the thunderstorm? take the speed of sound to be 339 m/s

Jobisdone [24]

Let the observer be 'd' distance away from the thunderstorm and let light take 't' time to reach the observer
Since the speed of sound and light remains constant in a particular medium, we can use
Speed = Distance/Time

For light,
3 x 10^8 = d/t
t = d/(3 x 10^8) -1

For sound,
339 = d/(t + 30) -2

Putting value from 1 in 2.
d = 10^4 m(approx)

3 0

3 years ago

In a thundercloud there may be an electric charge of +40 C near the top of the cloud and -40 C near the bottom of the cloud. The

Ratling [72]

Answer:

Electric force, $F=-3.59\times 10^6\ N$

Explanation:

Given that,

Electric charge 1, $q_1=+40\ C$

Electric charge 2, $q_2=-40\ C$

Distance, $d=2\ km=2\times 10^3\ m$

To find,

The electric force between these two sets of charges.

Solution,

There exists a force between two electric charges and this force is called electrostatic force. It is equal to the product of electric charged divided by square of distance between them.

$F=k\dfrac{q_1q_2}{d^2}$

k is the electrostatic constant

$F=8.99\times 10^9\times \dfrac{40\times (-40)}{(2\times 10^3)^2}$

$F=-3.59\times 10^6\ N$

So, the electric force between these two sets of charges is $-3.59\times 10^6\ N$ .

5 0

3 years ago

A particle travels clockwise on a circular path of diameter R, monitored by a sensor on the circle at point P; the other endpo

kotykmax [81]

We make a graphic of this problem to define the angle.

The angle we can calculate through triangle relation, that is,

$sin\theta = \frac{c}{QP}\\sin\theta = \frac{c}{R}\\\theta=sin^{-1}\frac{c}{R}$

With this function we should only calculate the derivate in function of c

$\frac{d\theta}{dc} = \frac{1}{\sqrt{1-\frac{c^2}{R^2}}}(\frac{c}{R})'\\\frac{d\theta}{dc} = \frac{1}{\sqrt{R^2-c^2}}$

That is the rate of change of $\theta$ .

b) At this point we need only make a substitution of 0 for c in the equation previously found.

$\frac{d\theta}{dc}\big|_{c=0} = \frac{1}{\sqrt{R^2-0}}\\\frac{d\theta}{dc}\big|_{c=0} = \frac{1}{R}$

Hence we have finally the rate of change when c=0.

6 0

2 years ago

A sound source A and a reflecting surface B move directly toward each other. Relative to the air, the speed of source A is 28.7

aleksandrvk [35]

(a) 1440.5 Hz

The general formula for the Doppler effect is

$f'=(\frac{v+v_r}{v+v_s})f$

where

f is the original frequency

f is the apparent frequency

$v$ is the velocity of the wave

$v_r$ is the velocity of the receiver (positive if the receiver is moving towards the source, negative otherwise)

$v_s$ is the velocity of the source (positive if the source is moving away from the receiver, negative otherwise)

Here we have

f = 1110 Hz

v = 334 m/s

In the reflector frame (= on surface B), we have also

$v_s = v_A = -28.7 m/s$ (surface A is the source, which is moving towards the receiver)

$v_r = +62.2 m/s$ (surface B is the receiver, which is moving towards the source)

So, the frequency observed in the reflector frame is

$f'=(\frac{334 m/s+62.2 m/s}{334 m/s-28.7 m/s})1110 Hz=1440.5 Hz$

(b) 0.232 m

The wavelength of a wave is given by

$\lambda=\frac{v}{f}$

where

v is the speed of the wave

f is the frequency

In the reflector frame,

f = 1440.5 Hz

So the wavelength is

$\lambda=\frac{334 m/s}{1440.5 Hz}=0.232 m$

(c) 1481.2 Hz

Again, we can use the same formula

$f'=(\frac{v+v_r}{v+v_s})f$

In the source frame (= on surface A), we have

$v_s = v_B = -62.2 m/s$ (surface B is now the source, since it reflects the wave, and it is moving towards the receiver)

$v_r = +28.7 m/s$ (surface A is now the receiver, which is moving towards the source)

So, the frequency observed in the source frame is

$f'=(\frac{334 m/s+28.7 m/s}{334 m/s-62.2 m/s})1110 Hz=1481.2 Hz$

(d) 0.225 m

The wavelength of the wave is given by

$\lambda=\frac{v}{f}$

where in this case we have

v = 334 m/s

f = 1481.2 Hz is the apparent in the source frame

So the wavelength is

$\lambda=\frac{334 m/s}{1481.2 Hz}=0.225 m$

8 0

3 years ago

How would the weight of a 7-kilogram concrete block compared to the weight of a 7-kilogram metal sphere?

Hitman42 [59]

Answer:

C. The block and the sphere would have the same weight.

Explanation:

If both the block and the sphere weigh 7 kilograms then they have the same weight. They may look different, but they both weigh 7 kilograms.

3 0

2 years ago

Read 2 more answers