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lord [1]
4 years ago
15

A woman is standing at the rim of a nonuniform cylindrical horizontal platform initially at rest. The platform is free to rotate

about frictionless orthogonal axle that goes through its center and has 4 m in diameter and moment of inertia of 500 kgm2. The woman then starts walking along the rim in clockwise direction at a constant speed of 1.50 m/s relative to the Earth. If a woman has 60 kg, how much work does she do to set herself and the platform into motion?
Physics
1 answer:
Natasha2012 [34]4 years ago
4 0

Answer:

e

Explanation:

i took it myself and got it right

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A firework is ignited, and explodes with a flash and a loud bang as it is blown apart. The system consists of: the firework, the
butalik [34]

The complete question is:

Study the scenario. A firework is ignited, and explodes with a flash and a loud bang as it is blown apart. The system consists of: the firework, the earth, and the air. which choice best describes how energy is transformed in the system?

A) When the firework is ignited, a chemical reaction absorbs energy from the surrounding environment. the energy is in several forms including sound and light, and mechanical energy of the fragments of the firecracker that are launched through the air. Eventually all the energy released is transformed into thermal energy.

B) When the firework is ignited, a chemical reaction releases energy in several forms, including sound, light, and the mechanical energy of the fragments being launched through the air. Eventually all the energy released is transformed into mechanical energy.

C) When the firework is ignited, a chemical reaction releases energy in several forms, including sound, light, and the mechanical energy of the fragments being launched through the air. Eventually all the energy released is transformed into thermal or mechanical energy.

D) When the firework is ignited, a chemical reaction absorbs energy from the surrounding environment. The energy is taken in in several forms including sound and light, and mechanical energy of the fragments being launched through the air. Eventually all the energy is transformed into thermal energy.

Answer:

C) When the firework is ignited, a chemical reaction releases energy in several forms, including sound, light, and the mechanical energy of the fragments being launched through the air. Eventually all the energy released is transformed into thermal or mechanical energy.

Explanation:

Energy is released from the system, not absorbed.

All the sound, light and movement of the debris is as a result of energy transformation from the chemical energy.

Eventually, most of the energy are finally wasted away as heat energy.

Some of the energy is used up by the flying particles from the fireworks.

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El peso de María es de 617,4 N. Si en dos meses pierde 2,5 Kg de masa, ¿cuál será su nuevo peso?
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Consider a cyclotron in which a beam of particles of positive charge q and mass m is moving along a circular path restricted by
Ulleksa [173]

A) v=\sqrt{\frac{2qV}{m}}

B) r=\frac{mv}{qB}

C) T=\frac{2\pi m}{qB}

D) \omega=\frac{qB}{m}

E) r=\frac{\sqrt{2mK}}{qB}

Explanation:

A)

When the particle is accelerated by a potential difference V, the change (decrease) in electric potential energy of the particle is given by:

\Delta U = qV

where

q is the charge of the particle (positive)

On the other hand, the change (increase) in the kinetic energy of the particle is (assuming it starts from rest):

\Delta K=\frac{1}{2}mv^2

where

m is the mass of the particle

v is its final speed

According to the law of conservation of energy, the change (decrease) in electric potential energy is equal to the increase in kinetic energy, so:

qV=\frac{1}{2}mv^2

And solving for v, we find the speed v at which the particle enters the cyclotron:

v=\sqrt{\frac{2qV}{m}}

B)

When the particle enters the region of magnetic field in the cyclotron, the magnetic force acting on the particle (acting perpendicular to the motion of the particle) is

F=qvB

where B is the strength of the magnetic field.

This force acts as centripetal force, so we can write:

F=m\frac{v^2}{r}

where r is the radius of the orbit.

Since the two forces are equal, we can equate them:

qvB=m\frac{v^2}{r}

And solving for r, we find the radius of the orbit:

r=\frac{mv}{qB} (1)

C)

The period of revolution of a particle in circular motion is the time taken by the particle to complete one revolution.

It can be calculated as the ratio between the length of the circumference (2\pi r) and the velocity of the particle (v):

T=\frac{2\pi r}{v} (2)

From eq.(1), we can rewrite the velocity of the particle as

v=\frac{qBr}{m}

Substituting into(2), we can rewrite the period of revolution of the particle as:

T=\frac{2\pi r}{(\frac{qBr}{m})}=\frac{2\pi m}{qB}

And we see that this period is indepedent on the velocity.

D)

The angular frequency of a particle in circular motion is related to the period by the formula

\omega=\frac{2\pi}{T} (3)

where T is the period.

The period has been found in part C:

T=\frac{2\pi m}{qB}

Therefore, substituting into (3), we find an expression for the angular frequency of motion:

\omega=\frac{2\pi}{(\frac{2\pi m}{qB})}=\frac{qB}{m}

And we see that also the angular frequency does not depend on the velocity.

E)

For this part, we use again the relationship found in part B:

v=\frac{qBr}{m}

which can be rewritten as

r=\frac{mv}{qB} (4)

The kinetic energy of the particle is written as

K=\frac{1}{2}mv^2

So, from this we can find another expression for the velocity:

v=\sqrt{\frac{2K}{m}}

And substitutin into (4), we find:

r=\frac{\sqrt{2mK}}{qB}

So, this is the radius of the cyclotron that we must have in order to accelerate the particles at a kinetic energy of K.

Note that for a cyclotron, the acceleration of the particles is achevied in the gap between the dees, where an electric field is applied (in fact, the magnetic field does zero work on the particle, so it does not provide acceleration).

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Do you believe in God (why or why not)?
masya89 [10]
Yes I believe in god.I believe in him because there is huge gaps in science that can not be solved. I also believe in him because I was also taught about him when I was little by family & my mom so I grew up learning about him which made my beliefs stronger of course.
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The power of a purely resistive lead is always positive although the current and voltage are sometimes negative. explain​
pickupchik [31]

Answer:

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