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3 years ago

14

Neutron stars are extremely dense objects that are formed from the remnants of supernova explosions. Many rotate very rapidly. S

uppose the mass of a certain spherical neutron star is twice the mass of the Sun and its radius is 13.0 km. Determine the greatest possible angular speed the neutron star can have so that the matter at its surface on the equator is just held in orbit by the gravitational force. (The mass of the Sun is 1.99 1030 kg.)

1 answer:

Vlada [557]3 years ago

5 0

Answer:

The required angular speed the neutron star is 10992.32 rad/s

Explanation:

Given the data in the question;

mass of the sun M $_S$ = 1.99 × 10³⁰ kg

Mass of the neutron star

M $_N$ = 2( M $_S$ )

M $_N$ = 2( 1.99 × 10³⁰ kg )

M $_N$ = ( 3.98 × 10³⁰ kg )

Radius of neutron star R $_N$ = 13.0 km = 13 × 10³ m

Now, let mass of a small object on the neutron star be m

angular speed be ω $_N$ .

During rotational motion, the gravitational force on the object supplies the necessary centripetal force.

GmM $_N$ = / R $_N$ ² = mR $_N$ ω $_N$ ²

ω $_N$ ² = GM $_N$ = / R $_N$ ³

ω $_N$ = √(GM $_N$ = / R $_N$ ³)

we know that gravitational G = 6.67 × 10⁻¹¹ Nm²/kg²

we substitute

ω $_N$ = √( ( 6.67 × 10⁻¹¹ )( 3.98 × 10³⁰ ) ) / (13 × 10³ )³)

ω $_N$ = √( 2.65466 × 10²⁰ / 2.197 × 10¹²

ω $_N$ = √ 120831133.3636777

ω $_N$ = 10992.32 rad/s

Therefore, The required angular speed the neutron star is 10992.32 rad/s

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geniusboy [140]

Answer:

a) the total distance traveled by the sphere during one cycle of its motion = 13.60 cm

b) The maximum speed is = 102.54 cm/s

The maximum speed occurs at maximum excursion from equilibrium.

c) The maximum magnitude of the acceleration of the sphere is = 30.93 $m/s^2$

The maximum acceleration occurs at maximum excursion from equilibrium.

Explanation:

Given that :

Frequency (f) = 4.80 Hz

Amplitude (A) = 3.40 cm

a)

The total distance traveled by the sphere during one cycle of simple harmonic motion is:

d = 4A (where A is the Amplitude)

d = 4(3.40 cm)

d = 13.60 cm

Hence, the total distance traveled by the sphere during one cycle of its motion = 13.60 cm

b)

As we all know that:

$x = Asin \omega t$

Differentiating the above expression with respect to x ; we have :

$\frac{d}{dt}(x) = \frac{d}{dt}(Asin \omega t)$

$v = A \omega cos \omega t$

Assuming the maximum value of the speed(v) takes place when cosine function is maximum and the maximum value for cosine function is 1 ;

Then:

$v_{max} = A \omega$

We can then say that the maximum speed therefore occurs at the mean (excursion) position where ; x = 0 i.e at maximum excursion from equilibrium

substituting $2 \pi f$ for $\omega$ in the above expression;

$v_{max} = A(2 \pi f)$

$v_{max} = 3.40 cm (2 \pi *4.80)$

$v_{max} = 102.54 \ cm/s$

Therefore, the maximum speed is = 102.54 cm/s

The maximum speed occurs at maximum excursion from equilibrium.

c) Again;

$v = A \omega cos \omega t$

By differentiation with respect to t;

$\frac{d}{dt}(v) = \frac{d}{dt}(A \omega cos \omega t)$

$a =- A \omega^2 sin \omega t$

The maximum acceleration of the sphere is;

$a_{max} =A \omega^2$

where;

$w = 2 \pi f$

$a_{max} = A(2 \pi f)^2$

where A= 3.40 cm = 0.034 m

$a_{max} = 0.034*(2 \pi *4.80)^2$

$a_{max} = 30.93 \ m/s^2$

The maximum magnitude of the acceleration of the sphere is = 30.93 $m/s^2$

The maximum acceleration occurs at maximum excursion from equilibrium where the oscillating sphere will have maximum acceleration at the turning points when the sphere has maximum displacement of $x = \pm A$

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3 years ago

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SashulF [63]

Answer:

3.28 degree

Explanation:

We are given that

Distance between the ruled lines on a diffraction grating, d=1900nm= $1900\times 10^{-9}m$

Where $1nm=10^{-9} m$

$\lambda_2=400nm=400\times10^{-9}m$

$\lambda_1=700nm=700\times 10^{-9}m$

We have to find the angular width of the gap between the first order spectrum and the second order spectrum.

We know that

$\theta=sin^{-1}(\frac{m\lambda}{d})$

Using the formula

m=1

$\theta_1=sin^{-1}(\frac{1\times700\times 10^{-9}}{1900\times 10^{-9}})$

$\theta=21.62^{\circ}$

Now, m=2

$\theta_2=sin^{-1}(\frac{2\times400\times 10^{-9}}{1900\times 10^{-9}})$

$\theta_2=24.90^{\circ}$

$\Delta \theta=\theta_2-\theta_1$

$\Delta \theta=24.90-21.62$

$\Delta \theta=3.28^{\circ}$

Hence, the angular width of the gap between the first order spectrum and the second order spectrum=3.28 degree

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v^2=v(original)+2a*(change in x)
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Then using v(avg)=x/t
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nydimaria [60]

Answer:

Explanation:

The formula for this is

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Filling in:

$7.5*10^{-8}=\frac{6.67*10^{-11}(90.0)(30.0)}{r^2}$ and moving things around to solve for r:

$r=\sqrt{\frac{6.67*10^{-11}(90.0)(30.0)}{7.5*10^{-8}} }$ Doing all that and rounding to the 3 sig fig's you need gives us a distance of 1.55 m

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