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allochka39001 [22]
3 years ago
7

A copper wire has a circular cross section with a radius of 1.00 mm. (a) If the wire carries a current of 2.80 A, find the drift

speed of the electrons in the wire. (Assume the density of charge carriers (electrons) in a copper wire is n = 8.46 1028 electrons/m3.) 2.1e-6 Incorrect: Your answer is incorrect. Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. m/s (b) All other things being equal, what happens to the drift speed in wires made of metal having a larger number of conduction electrons per atom than copper? Explain

Physics
1 answer:
Sonja [21]3 years ago
7 0

Answer:

Vd = 6.58×10-⁵m/s

Explanation:

See the attachment below.

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The composition of the protoplanetary disk varies with distance from the protostar due to temperature.
lesantik [10]

Answer:

b) 1. Iron, silicates, carbon

   2. Water

   3. Methane, Ammonia, Carbon Dioxide.

Explanation:

Protoplanetry disk is the disk of gases and clouds of dust that rotates around the newly made star. The temperature of the protoplanetry disk actually determines the type of the planet that is to be formed. Inner part of the protoplanetry disk is closer to the sun thats why it is the hottest and denser part and composed of the materials like Iron, silicates, carbon as they have high melting points. Then comes those materials that exist in the solid form at lower temperatures such as the volatile materials like water. Ater that the protoplanetry disk is made of highly volatile materials that exists in solid from only at low coldest temperatures. So the outer part of the protoplanetry disk  is made up of the Methane, Ammonia and Carbon Dioxide.

7 0
3 years ago
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).

6 0
3 years ago
Which of the following must be true for “P if and only if Q” to be true?
erastovalidia [21]
Explain or message me what your trying to ask!

6 0
3 years ago
PLEASE !! Help
n200080 [17]

Answer:

scalar,magnitude

Explanation:

scalar is an example of length/distance

magnitude is the length of a vector

8 0
2 years ago
An electric car uses a 45-kW (160-hp) motor. If the battery pack is designed for 340V, what current would the motor need to draw
alukav5142 [94]

Answer:

Current = 132.35 A

The motor needs to draw 132.35 Amperes current from the battery.

Explanation:

The formula of electric power is given as follows:

Power = (Voltage)(Current)

Current = Power/Voltage

In this question, we have:

Power = 45 KW = 45000 W

Voltage of Battery Pack = 340 V

Current needed to be drawn = ?

Therefore,

Current = 45000 W/340 V

<u>Current = 132.35 A</u>

<u>The motor needs to draw 132.35 Amperes current from the battery.</u>

5 0
2 years ago
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