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horsena [70]
3 years ago
10

Drift speed refers to the average distance traveled by a charge carrier per unit of time. Like the speed of any object, the drif

t speed of an electron moving through a wire is the distance to time ratio. A typical drift speed might be 1 meter per hour. That is slow! 1. If the average drift speed of an electron is very slow, how can there by a current on the order of 1 or 2 ampere in
Physics
1 answer:
vovikov84 [41]3 years ago
5 0

Answer:

Explanation:

there are many, many charge carriers moving directly throughout the entire length of the circuit. Current is that the rate at which charge crosses some extent on a circuit. A high current is that the results of several coulombs of charge crossover a cross section of a wire on a circuit. If the charge carriers are densely packed into the wire, then there doesn't need to be a high speed to possess a high current. That is, the charge carriers don't need to travel an extended distance during a second, there just has got to be tons of them passing through the cross section.

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1. What is the velocity of an 250-kg that has 6,250 J of energy?<br><br><br> Please help
leonid [27]

Explanation:

KE = 1/2mv^2

6,250 J =1/2(250kg) v ^2

<u>6250 J </u>= <u>125kgv^2</u>

125kg. 125kg

v =

\sqrt{50}

6 0
3 years ago
A small uniform disk and a small uniform sphere are released simultaneously at the top of a high inclined plane, and they roll d
barxatty [35]

Answer:

(D) the sphere

Explanation:

The bodies given are Disk and Solid sphere (uniform sphere)

Moment of inertia of the bodies are

I(disk) = \frac{MR^2}{2}

I(sphere) = \frac{2MR^2}{5}

Since the moment of inertia of sphere is less than that of disk, therefore sphere will reach the bottom first.

3 0
3 years ago
A solenoid used to produce magnetic fields for research purposes is 2.1 m long, with an inner radius of 28 cm and 1000 turns of
Veseljchak [2.6K]

Answer:

I = 2172.46 A

Explanation:

Given that,

The length of a solenoid, l = 2.1 m

The inner radius of the solenoid, r = 28 cm = 0.28 m

The number of turns in the wire, N = 1000

The magnetic field in the solenoid, B = 1.3 T

We need to find the current carried by it. We know that, the magnetic field in a solenoid is given by :

B=\mu_o nI\\\\or\\\\B=\mu_o \dfrac{N}{L}I\\\\I=\dfrac{BL}{\mu_o N}

Put all the values,

I=\dfrac{1.3\times 2.1}{4\pi \times 10^{-7}\times 1000}\\\\I=2172.46\ A

So, it carry current of 2172.46 A.

7 0
3 years ago
An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km fr
mestny [16]

An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km from its center. The black hole is 5.00 times the mass of the sun and has a Schwarzschild radius of 15.0 km. The astronaut is positioned inside the spaceship such that one of her 0.030 kg ears is 6.0 cm farther from the black hole than the center of mass of the spacecraft and the other ear is 6.0 cm closer.

What is the tension between her ears?

Would the astronaut find it difficult to keep from being torn apart by the gravitational forces?

Answer:

The tension between the ears = 2.07 KN

The astronaut will find it difficult to keep and will eventually be in trouble because the tension is now greater compared to the tension in the human tissues.

Explanation:

Given that:

Orbital radius of the spacecraft (R) = 120 Km = 120 × 10³ m

Mass of the black hole (m) = 5 \ * (M \ _{sun})

where : M_{sun} = 1.99*10^{33} \ kg

Then; we have:

 m = 5*(1.99*10^{30} \ kg ) \\ = 9.95*10^{30} kg

Schwarzchild radius of the black hole

r - 15.0 km

Mass of each ear m_{ear} = 0.030 \ kg

Farther distance between one ear and the black hole (d) = 6.0 cm

= 0.06 m

Closer distance between the other ear and the black home is (d) 6.0 cm

= 0.6 cm

NOW, If we assume that the tension force should be T; then definitely the two ears will posses the same angular velocity .

The net force on the ear closer to the black hole will be:

\frac{GMm_{ear} }{(R-d)}- T = m_{ear} (R -  d) \omega^2

\frac{GMm_{ear} }{(R-d)^2}- \frac{T}{(R-d)} = m_{ear} \omega^2 \ ----> \ (1)

The net force on the ear farther to the black hole is :

\frac{GMm_{ear} }{(R+d)}- T = m_{ear} (R +  d) \omega^2

\frac{GMm_{ear} }{(R+d)^2}- \frac{T}{(R+d)} = m_{ear} \omega^2 \ ----> \ (2)

Equating equation (1) and (2) & therefore making (T) the subject of the formula; we have:

T = \frac{3GMm_{ear}d}{R^3}

T = \frac{3(6.67*10^{-11}N.m^2/kg^2)(1.95*10^{30}kg)(0.03kg)(0.06m)}{(120*10^3m)^3}

T = 2073.9 N\\T = 2.07 KN

The tension between the ears = 2.07 KN

The astronaut will find it difficult to keep and will eventually be in trouble because the tension is now greater compared to the tension in the human tissues.

3 0
4 years ago
For the potential , determine the number of bound states?A)6 b)4 c) 3 d) 1
s2008m [1.1K]

Do you see that blank, open space after the word "potential ..." ?

There's supposed to be a number there that actually tells us the value of the potential.  Without that number ... and a lot more description of the whole scenario here ... there's no possible answer to the question.

7 0
3 years ago
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