3. Which object has more inertia?A. A tractor trailer rig moving at 2 m/s
B. A pingpong ball rolling a 2 m/s
C. A bowling ball rolling at 1m/s
D. A car rolling at 5 m/s

Answers

Answer 1

Answer:

A. A tractor trailer rig moving at 2 m/s

Explanation:

Inertia can be defined as the tendency of an object or a body to continue in its state of motion or remain at rest unless acted upon by an external force.

In physics, Sir Isaac Newton's first law of motion is known as law of inertia and it states that, an object or a physical body in motion will continue in its state of motion at continuous velocity (the same speed and direction) or, if at rest, will remain at rest unless acted upon by an external force.

The inertia of an object such as a tractor trailer rig is greatly dependent or influenced by its mass; the higher quantity of matter in a tractor trailer rig, the greater will be its tendency to continuously remain at rest.

Hence, the object that has more inertia is a tractor trailer rig moving at 2 m/s because it has more mass than all the other objects in the category. Also, the mass of an object is directly proportional to its inertia.

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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:

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

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:

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Halogen lightbulbs allow their filaments to operate at a higher temperature than the filaments in standard incandescent bulbs. For comparison, the filament in a standard lightbulb operates at about 2900K, whereas the filament in a halogen bulb may operate at 3400K. Which bulb has the higher peak frequency? Calculate the ratio of the peak frequencies. The human eye is most sensitive to a frequency around 5.5x10^14 Hz. Which bulb produces a peak frequency close to this value?

Answers

Answer:

Halogen

0.85294

Explanation:

c = Speed of light = $3* 10^8\ m/s$

b = Wien's displacement constant = $2.897* 10^(-3)\ mK$

T = Temperature

From Wien's law we have

$\lambda_m=(b)/(T)\n\Rightarrow \lambda_m=(2.897* 10^(-3))/(2900)\n\Rightarrow \lambda_m=9.98966* 10^(-7)\ m$

Frequency is given by

$\nu=(c)/(\lambda_m)\n\Rightarrow \nu=(3* 10^8)/(9.98966* 10^(-7))\n\Rightarrow \nu=3.00311* 10^(14)\ Hz$

For Halogen

$\lambda_m=(b)/(T)\n\Rightarrow \lambda_m=(2.897* 10^(-3))/(3400)\n\Rightarrow \lambda_m=8.52059* 10^(-7)\ m$

Frequency is given by

$\nu=(c)/(\lambda_m)\n\Rightarrow \nu=(3* 10^8)/(8.52059* 10^(-7))\n\Rightarrow \nu=3.52088* 10^(14)\ Hz$

The maximum frequency is produced by Halogen bulbs which is closest to the value of $5.5* 10^(14)\ Hz$

Ratio

$(3.00311* 10^(14))/(3.52088* 10^(14))=0.85294$

The ratio of Incandescent to halogen peak frequency is 0.85294

Air enters an adiabatic compressor at 104 kPa and 292 K and exits at a temperature of 565 K. Determine the power (kW) for the compressor if the inlet volumetric flow rate is 0.15 m3/s. Use constant specific heats evaluated at 300 K.

Answers

Answer:

$\dot W_(in) = 49.386\,kW$

Explanation:

An adiabatic compressor is modelled as follows by using the First Law of Thermodynamics:

$\dot W_(in) + \dot m \cdot c_(p)\cdot (T_(1)-T_(2)) = 0$

The power consumed by the compressor can be calculated by the following expression:

$\dot W_(in) = \dot m \cdot c_(v)\cdot (T_(2)-T_(1))$

Let consider that air behaves ideally. The density of air at inlet is:

$P\cdot V = n\cdot R_(u)\cdot T$

$P\cdot V = (m)/(M)\cdot R_(u)\cdot T$

$\rho = (P\cdot M)/(R_(u)\cdot T)$

$\rho = ((104\,kPa)\cdot (28.02\,(kg)/(kmol)))/((8.315\,(kPa\cdot m^(3))/(kmol\cdot K) )\cdot (292\,K))$

$\rho = 1.2\,(kg)/(m^(3))$

The mass flow through compressor is:

$\dot m = \rho \cdot \dot V$

$\dot m = (1.2\,(kg)/(m^(3)))\cdot (0.15\,(m^(3))/(s) )$

$\dot m = 0.18\,(kg)/(s)$

The work input is:

$\dot W_(in) = (0.18\,(kg)/(s) )\cdot (1.005\,(kJ)/(kg\cdot K))\cdot (565\,K-292\,K)$

$\dot W_(in) = 49.386\,kW$

A conducting loop of radius r=0.1 m, carrying a current I=2 A has a magnetic moment \vec{\mu} μ that is entirely in the j-hat direction. The loop is immersed in a magnetic field \vec{B} B = [3 i-hat + 4 j-hat] T. What is the potential energy of the loop in this configuration?

Answers

Answer:

Explanation:

Magnetic moment of current carrying loop

= current x area

= 2 x π x .1²

M = .0628 unit . it is in j direction so vecor form of it

M = .0628 j

Magnetic field B = 3i + 4 j

Energy

= - M.B

- .0628 j . ( 3i + 4 j )

= - .2512 J

Which of the following best represents stored potential energy?Air leaking from a flat tire
Stress built up in a rock fault
Heat given off by a forest fire
Water flowing through a hose

Answers

Answer:

Explanation:

stress built up on a rock fault

A rectangular loop (area = 0.15 m2) turns in a uniform magnetic field, B = 0.18 T. When the angle between the field and the normal to the plane of the loop is π/2 rad and increasing at 0.75 rad/s, what emf is induced in the loop?

Answers

Answer:

Emf induced in the loop is 0.02V

Explanation:

To get the emf of induced loop, we have to use faraday's law

ε = - dΦ/dt

To get the flux, we use;

Φ = BA cos(θ)

B = The uniform magnetic field

A = Area of rectangular loop

θ = angle between magnetic field and normal to the plane of loop

substitute the flux equation (Φ) into the faraday's equation

we have ε = - d(BA cos(θ)) / dt

ε = BA sinθ dθ/dt

from the question;B = 0.18T, A=0.15m2, θ = π/2 ,dθ/dt = 0.75rad/s

Our equation will now look like this;

ε = (0.18T) (0.15m2) (sin(π/2)) (0.75rad/s)

ε = 0.02V

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3. Which object has more inertia?A. A tractor trailer rig moving at 2 m/sB. A pingpong ball rolling a 2 m/sC. A bowling ball rolling at 1m/sD. A car rolling at 5 m/s

Answers

Related Questions

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Final answer:

Explanation:

Learn more about Angular velocity of a supernova shell here:

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3. Which object has more inertia?A. A tractor trailer rig moving at 2 m/s
B. A pingpong ball rolling a 2 m/s
C. A bowling ball rolling at 1m/s
D. A car rolling at 5 m/s