PHYSICS COLLEGE

When some stars use up their fuel, they undergo a catastrophic explosion called a supernova. This explosion blows much or all of a star's mass outward, in the form of a rapidly expanding spherical shell. As a simple model of the supernova process, assume that the star is a solid sphere of radius R that is initially rotating at 2.4 revolutions per day. After the star explodes, find the angular velocity, in revolutions per day, of the expanding supernova shell when its radius is 4.3R. Assume that all of the star's original mass is contained in the shell.

Answers

Answer 1

Answer:

Answer:

0.0768 revolutions per day

Explanation:

R = Radius

$\omega$ = Angular velocity

As the mass is conserved the angular momentum is conserved

$I_1\omega_1=I_2\omega_2\n\Rightarrow (I_1)/(I_2)=(\omega_2)/(\omega_1)$

Moment of intertia for solid sphere

$I_1=(2)/(5)MR^2\n\Rightarrow I_1=0.4MR^2$

Moment of intertia for hollow sphere

$I_2=(2)/(3)M(4.3R)^2\n\Rightarrow I_2=12.327MR^2$

Dividing the moment of inertia

$(I_1)/(I_1)=(0.4MR^2)/(12.327MR^2)\n\Rightarrow (I_1)/(I_2)=0.032$

From the first equation

$\omega_2=\omega_1(I_1)/(I_2)\n\Rightarrow \omega_2=2.4* 0.032\n\Rightarrow \omega_2=0.0768\ rev\day$

The angular velocity, in revolutions per day, of the expanding supernova shell is 0.0768 revolutions per day

Answer 2

Answer:

Final answer:

To find the angular velocity of the expanding supernova shell, we can use the principle of conservation of angular momentum. The initial angular momentum of the star can be equated to the final angular momentum of the shell. By substituting the given information and solving the equation, we can find the angular velocity of the shell.

Explanation:

When a star undergoes a supernova explosion, a large amount of its mass is blown outward in the form of a rapidly expanding shell. To find the angular velocity of the expanding shell, we can use the principle of conservation of angular momentum. Assuming that all of the star's original mass is contained in the shell, we can equate the initial angular momentum of the star to the final angular momentum of the shell.

The angular velocity of the star before the explosion can be calculated using the equation:

angular velocity before = 2 * pi * initial frequency

where the initial frequency is given as 2.4 revolutions per day.

After the explosion, the radius of the expanding shell is given as 4.3 times the radius of the star. Using the principle of conservation of angular momentum, we can set the initial angular momentum of the star equal to the final angular momentum of the shell:

initial angular momentum of the star = final angular momentum of the shell

Since the final angular momentum of the shell is given by:

final angular momentum of the shell = moment of inertia of the shell * angular velocity of the shell

where the moment of inertia of the shell is given by:

moment of inertia of the shell = 2/5 * mass of the shell * (radius of the shell)^2

and the angular velocity of the shell is what we are trying to find, we can rewrite the equation as:

initial angular momentum of the star = 2/5 * mass of the shell * (radius of the shell)^2 * angular velocity of the shell

By substituting the expression for the initial angular momentum of the star and solving for the angular velocity of the shell, we can find the answer.

Learn more about Angular velocity of a supernova shell here:

brainly.com/question/13896592

#SPJ11

Related Questions

A system consists of two particles. The first particle has mass m1 = 6.60 kg and a velocity of (4.20i + 2.00j) m/s, and the second particle has mass m2 = 2.00 kg and a velocity of (2.00i − 3.60j) m/s. (Express your answers in vector form.)Required:a. What is the velocity of the center of mass of this system?b. What is the total momentum of this system?
The fundamental force that is responsible for beta decay is the: A) weak force. B) strong force. C) dark energy force. D) gravitational force E) electromagnetic force.
An object having a net charge of 25.0 µC is placed in a uniform electric field of 620 N/C directed vertically. What is the mass of this object if it "floats" in the field?
The switch in the circuit has been in the left position for a long time. At t=0 it moves to the right position and stays there.a. Write the expression for the capacitor voltage v(t), fort≥0. b. Write the expression for the current through the 2.4kΩ resistor, i(t), fort≥0+.
According to the Heisenberg uncertainty principle, quantum mechanics differs from classical mechanics in that: Select the correct answer below: Quantum mechanics involves particles that do not move. It is impossible to calculate with accuracy both the position and momentum of particles in classical mechanics. The measurement of an observable quantity in the quantum domain inherently changes the value of that quantity. All of the above

The tensile strength (the maximum tensile stress it can support without breaking) for a certain steel wire is 3000 MN/m2. What is the maximum load that can be applied to a wire with a diameter of 3.0 mm made of this steel without breaking the wire?

Answers

Answer:

The correct answer is "21195 N".

Explanation:

The given values are:

Tensile strength,

= 3000 MN/m²

Diameter,

= 3.0 mm

i.e.,

= 3×10⁻³ m

Now,

The maximum load will be:

= $Tensile \ strength* Area$

On substituting the values, we get

= $(3000* 10^6)((\pi)/(4) (3* 10^(-3))^2)$

= $(3000* 10^6)((3.14)/(4) (3* 10^(-3))^2)$

= $21195 \ N$

Final answer:

The maximum load that can be applied to a 3.0 mm diameter steel wire with a tensile strength of 3000 MN/m2 without breaking it is 21,200 Newtons.

Explanation:

The subject of this question revolves around the concept of tensile strength in the field of Physics. The maximum load that can be applied to a wire without it breaking depends on the wire's tensile strength and its cross-sectional area. For a steel wire with a tensile strength of 3000 MN/m2 and a diameter of 3.0 mm, we first need to calculate the cross-sectional area, which can be found using the formula for the area of a circle, A = πr^2, where r is the radius of the wire. Given the diameter is 3.0 mm, the radius will be 1.5 mm or 1.5 x 10^-3 m. So, A = π(1.5 x 10^-3 m)^2 ≈ 7.07 x 10^-6 m^2.

We can then use the tensile strength (σ) to find the maximum load (F) using the equation F = σA. Substituting the given values, we get F = 3000 MN/m^2 * 7.07 x 10^-6 m^2 = 21.2 kN, which is equivalent to 21,200 N. Therefore, the maximum load that can be applied to the wire without breaking it is 21,200 Newtons.

Learn more about Tensile Strength here:

brainly.com/question/14293634

#SPJ3

The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in meters, t in seconds, and c and b are positive constants. What are the units of (a) constant c and (b) constant b? Find a formula in terms of c, b, and t of the (c) velocity v and (d) acceleration a. (e) At what time t ≥ 0 does the particle reach its maximum x value?

Answers

Answer:

(a): $\rm meter/ second^2.$

(b): $\rm meter/ second^3.$

(c): $\rm 2ct-3bt^2.$

(d): $\rm 2c-6bt.$

(e): $\rm t=(2c)/(3b).$

Explanation:

Given, the position of the particle along the x axis is

$\rm x=ct^2-bt^3.$

The units of terms $\rm ct^2$ and $\rm bt^3$ should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of $\rm ct^2=meter$

Therefore, unit of $\rm c= meter/ second^2.$

(b):

Unit of $\rm bt^3=meter$

Therefore, unit of $\rm b= meter/ second^3.$

(c):

The velocity v and the position x of a particle are related as

$\rm v=(dx)/(dt)\n=(d)/(dx)(ct^2-bt^3)\n=2ct-3bt^2.$

(d):

The acceleration a and the velocity v of the particle is related as

$\rm a = (dv)/(dt)\n=(d)/(dt)(2ct-3bt^2)\n=2c-6bt.$

(e):

The particle attains maximum x at, let's say, $\rm t_o$ , when the following two conditions are fulfilled:

$\rm \left ((dx)/(dt)\right )_(t=t_o)=0.$
$\rm \left ( (d^2x)/(dt^2)\right )_(t=t_o)<0.$

Applying both these conditions,

$\rm \left ( (dx)/(dt)\right )_(t=t_o)=0\n2ct_o-3bt_o^2=0\nt_o(2c-3bt_o)=0\nt_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = (2c)/(3b).$

For $\rm t_o = 0$ ,

$\rm \left ( (d^2x)/(dt^2)\right )_(t=t_o)=2c-6bt_o = 2c-6\cdot 0=2c$

Since, c is a positive constant therefore, for $\rm t_o = 0$ ,

$\rm \left ( (d^2x)/(dt^2)\right )_(t=t_o)>0$

Thus, particle does not reach its maximum value at $\rm t = 0\ s.$

For $\rm t_o = (2c)/(3b)$ ,

$\rm \left ( (d^2x)/(dt^2)\right )_(t=t_o)=2c-6bt_o = 2c-6b\cdot (2c)/(3b)=2c-4c=-2c.$

Here,

$\rm \left ( (d^2x)/(dt^2)\right )_(t=t_o)<0.$

Thus, the particle reach its maximum x value at time $\rm t_o = (2c)/(3b).$

At the normal boiling temperature of iron, TB = 3330 K, the rate of change of the vapor pressure of liquid iron with temperature is 3.72 x 10-3 atm/K. Calculate the molar latent enthalpy of boiling of iron at 3330 K:

Answers

The molar latent enthalpy of boiling of iron at 3330 K is ΔH = 342 $*$ 10^3 J.

Explanation:

Molar enthalpy of fusion is the amount of energy needed to change one mole of a substance from the solid phase to the liquid phase at constant temperature and pressure.

d ln p = (ΔH / RT^2) dt

(1/p) dp = (ΔH / RT^2) dt

dp / dt = p (ΔH / RT^2) = 3.72 $*$ 10^-3

(p) (ΔH) / (8.31) (3330)^2 = 3.72 $*$ 10^-3

ΔH = 342 $*$ 10^3 J.

An electric current in a conductor varies with time according to the expression i(t) = 110 sin (120πt), where i is in amperes and t is in seconds. what is the total charge passing a given point in the conductor from t = 0 to t = 1/180 s?

Answers

As we know that current is defined as rate of flow of charge

$i = (dq)/(dt)$

so by rearranging the equation we can say

$q = \int i dt$

here we know that

$i(t) = 110 sin(120\pi t)$

here we will substitute it in the above equation

$q = \int 110 sin(120\pi t) dt$

$q = 110 [- (cos(120\pi t))/(120\pi)]$

now here limits of time is from t = 0 to t = 1/180s

so here it will be given as

$q = (110)/(120\pi)( -cos0 + cos((2\pi)/(3)))$

$q = 0.44 C$

so total charge flow will be 0.44 C

Answer:

The total charge passing a given point in the conductor is 0.438 C.

Explanation:

Given that,

The expression of current is

$i(t)=110\sin(120\pi t)$

$(dq(t))/(t)=110\sin(120\pi t)$

$dq(t)=110\sin(120\pi t)dt$ ....(I)

We need to calculate the total charge

On integrating both side of equation (I)

$\int_(0)^(q)dq(t)=\int_(0)^{(1)/(180)}110\sin(120\pi t)dt$

$q=110((-\cos(120\pi t))/(120\pi))_(0)^{(1)/(180)}$

$q=-(110)/(120\pi)(cos(120\pi((1)/(180)))-\cos120\pi(0))$

$q=-0.2918(-(1)/(2)-1)$

$q=0.438\ C$

Hence, The total charge passing a given point in the conductor is 0.438 C.

Please help really easy

Answers

Answer:

I am sure it is A because no chemical change occurs and it is a physical change. If you can Brainllest than that would be great but if you wanna you don't have to. Hope this helps!! If wrong sorry.

Explanation:

Light bulb 1 operates with a filament temperature of 2700 K whereas light bulb 2 has a filament temperature of 2100 K. Both filaments have the same emissivity, and both bulbs radiate the same power. Find the ratio A1/A2 of the filament areas of the bulbs.