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dalvyx [7]
3 years ago
13

(28 points) In a little over 5 billion years, our star will slough off ~20% of its mass and collapse to a white dwarf star of ra

dius 8,000 km. We will model it as a sphere. What will its angular momentum be in terms of its current angular momentum? What will its rotation period be in terms of its current period? b) (12 points) What will its rotational kinetic energy be in terms of its current rotational kinetic energy?
Physics
1 answer:
Sedbober [7]3 years ago
8 0

Answer:

The angular momentum is same as it was before.

The rotation period is 1.058*10^{-4} times the original period.

The rotational kinetic energy is 9452 times greater.

Explanation:

The angular momentum L of a rigid body is

L = I\omega,

where I is the moment of inertia and \omega is the angular velocity.

Now, the law of conservation of momentum demands that

I_1\omega_1 = I_2\omega_2,

in words this means the angular momentum before must equal the angular momentum after.

Let us call M the mass, R the radius, and \omega_1 the angular velocity of the sun before it becomes a white dwarf, then its linear momentum is

I_1\omega_1 = \dfrac{2}{5}MR^2 \omega_1     (Remember for a solid sphere I = \dfrac{2}{5} MR^2)

After it has become a white dwarf, the suns mass is 80% of what it had before <em>(went off by 20%),</em> and its radius has become 0.0115% its initial value <em>(8000 km is 0.0115% of the radius of the sun ); </em>therefore, the angular momentum is

I_2\omega_2 = \dfrac{2}{5} (0.8M)(0.0115R)^2 \omega_2

which must be equal to the angular momentum it had before; therefore

\dfrac{2}{5}MR^2 \omega_1 = \dfrac{2}{5} (0.8M)(0.0115R)^2 \omega_2

which we solve for \omega_2:

MR^2 \omega_1 = (0.8M)(0.0115R)^2 \omega_2

MR^2 \omega_1 = (0.8)(0.0115)^2 MR^2\omega_2

\omega_1 = (0.8)(0.0115)^2 \omega_2

\omega_2= \dfrac{\omega_1}{(0.8)(0.0115)^2 }

\boxed{ \omega_2 = 9452\omega_1.}

which is about whopping 9500 times larger than initial angular velocity!!

Now the rotation period T is

T_2 = \dfrac{2\pi}{\omega_2}

T = \dfrac{2\pi}{ 9452\omega_1}= 1.058*10^{-4} (\dfrac{2\pi}{ \omega_1})

since \dfrac{2\pi}{ \omega_1} =T_1

\boxed{T_2 = 1.058*10^{-4} T_1}

Similarly, the rotation kinetic energy will be

K_2 = \dfrac{1}{2}I_2\omega_2^2

K_2 = \dfrac{1}{2}*\dfrac{2}{5} (0.8M)(0.0115R)^2 ( 9452\omega_1})^2

K_2 =0.8*0.0115^2*9452^2 [\dfrac{1}{2}*\dfrac{2}{5} mR^2w_1^2]

\boxed{K_2 =9452 K_1}

which is about 9500 times larger than initial rotational kinetic energy!

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Jlenok [28]

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5 0
3 years ago
Toon Train is traveling at the speed of 10 m/s at the top of a hill. Five seconds later it reaches the bottom of the hill and is
Naddika [18.5K]

Answer:

the rate of acceleration of the train is 4 m/s²

Explanation:

Given;

initial velocity of the train, u = 10 m/s

change in time of motion, dt = 5 s

final velocity of the train, v = 30 m/s

The rate of acceleration of the train is calculated as;

a = \frac{dv}{dt} = \frac{v-u}{dt} = \frac{30-10}{5} = \frac{20}{5} = 4 \ m/s^2

Therefore, the rate of acceleration of the train is 4 m/s²

5 0
3 years ago
Explain why the "X" chromosome expresses more traits than the "y" chromosome.
wel

Explanation:

The X chromosome (females have 2 of them while males have 1) is five times larger than the Y chromosome and has 10 times the number of genes.

5 0
3 years ago
A proton is initially moving west at a speed of 1.10 106 m/s in a uniform magnetic field of magnitude 0.281 T directed verticall
kifflom [539]

Answer:

so here it will move in circle with radius 4.06 cm

Explanation:

As we know that proton is moving towards west while the magnetic field is vertically upwards

So here the force on the proton must be perpendicular to the velocity

So here we have

F = q(\vec v \times \vec B)

so here we have

F = qvB sin90

since force is perpendicular to the velocity so here it must be centripetal force

here we have

\frac{mv^2}{R} = qvB

so we have

R = \frac{mv}{qB}

R = \frac{(1.66 \times 10^{-27})(1.10 \times 10^6)}{(1.6 \times 10^{-19})(0.281)}

R = 4.06 cm

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8 0
3 years ago
Write the realtionship between the density of a liquid and its upthrust? clarify​
kipiarov [429]

Answer:

B = ρ g V_liquid

the thrust is proportional to the density of the liquid

Explanation:

The density of a liquid is defined as the relationship between the mass and the volume of the liquid

         ρ = m / V

The upward push of the liquid is given by the principle of Archimedes Archimedes establishes that the push is equal to the weight of the dislodged liquid

        B = W_liquid

        B = m _liquid g

we substitute mass for density

        B = ρ g V_liquid

therefore we see that the thrust is proportional to the density of the liquid

8 0
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