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Papessa [141]
3 years ago
6

The magnetic field strength at the north pole of a 2.0-cmcm-diameter, 8-cmcm-long Alnico magnet is 0.10 TT. To produce the same

field with a solenoid of the same size, carrying a current of 2.4 AA , how many turns of wire would you need
Physics
1 answer:
horrorfan [7]3 years ago
6 0

Answer:

2653 turns

Explanation:

We are given that

Diameter,d=2 cm

Length of magnet,l=8 cm=8\times 10^{-2} m

1m=100 cm

Magnetic field,B=0.1 T

Current,I=2.4 A

We are given that

Magnetic field of solenoid and magnetic are same and size of both solenoid and magnetic are also same.

Length of solenoid=8\times 10^{-2} m

Magnetic field of solenoid

B=\frac{\mu_0NI}{l}

Using the formula

0.1=\frac{4\pi\times 10^{-7}\times 2.4\times N}{8\times 10^{-2}}

Where \mu_0=4\pi\times 10^{-7}

N=\frac{0.1\times 8\times 10^{-2}}{4\pi\times 10^{-7}\times 2.4}=2653 turns

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Pressure, V is volume , T is temperature of a gas and R is gas constant.
Ilya [14]
هاذغهعفذعذهفذهذهغذهفهذهفذهف
6 0
3 years ago
What happens to the particles of a liquid when energy is removed from them?
KonstantinChe [14]

Answer:

D: The distance between the particles decreases

Explanation:

Taking away energy slows down molecules, like how you slow down when you are cold (I think)

3 0
4 years ago
A rectangular loop of area A is placed in a region where the magnetic field is perpendicular to the plane of the loop. The magni
mina [271]

Answer:

Induced emf, \epsilon=-A\dfrac{B_{max}e^{-t/\tau}}{\tau}

Explanation:

The varying magnetic field with time t is given by according to equation as :

B=B_{max}e^{-t/\tau}

Where

B_{max}\ and\ t are constant

Let \epsilon is the emf induced in the loop as a function of time. We know that the rate of change of magnetic flux is equal to the induced emf as:

\epsilon=-\dfrac{d\phi}{dt}

\epsilon=-\dfrac{d(BA)}{dt}

\epsilon=-A\dfrac{d(B)}{dt}

\epsilon=-A\dfrac{d(B_{max}e^{-t/\tau})}{dt}

\epsilon=A\dfrac{B_{max}e^{-t/\tau}}{\tau}

So, the induced emf in the loop as a function of time is A\dfrac{B_{max}e^{-t/\tau}}{\tau}. Hence, this is the required solution.

7 0
3 years ago
WOULUJUTUL RECIPECUIUS.
3241004551 [841]

The force between the two objects is 19.73 nN.

<u>Explanation: </u>

Any force acting between two objects tends to be directly proportional to the product of their masses and inversely proportional to the square of the distance between the two objects. And this kind of attraction force between two objects is termed as gravitational force.

So if we consider M_{1} and M_{2} as the masses of both objects and let d be the distance of separation of two objects. Then the force between the two objects can be determined as below:

                      \text {Gravitational force}=\frac{G \times M_{1} \times M_{2}}{d^{2}}

As gravitational constant G=6.67 \times 10^{-11} \mathrm{m}^{3} \mathrm{kg}^{-1} \mathrm{s}^{-2}, M_{1} = 20 kg and  M_{2} = 100 kg, while d = 2.6 m, then

                    \text {Gravitational force}=\frac{6.67 \times 10^{-11} \times 20 \times 100}{(2.6)^{2}}=\frac{6.67 \times 20 \times 10^{-9}}{6.76}

Thus, we get finally,

                   \text {Gravitational force}=19.73 \times 10^{-9} \mathrm{N}

As we know, nano denoted by letter 'n' equals to 10^{-9}

So the force acting between two objects is 19.73 nN.

7 0
3 years ago
A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begin
Mashutka [201]

The given question is incomplete. The complete question is as follows.

A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begins to move. Then the force (in newtons) exerted by the expanding gas on the bullet is 14,000 + 10,000x − 26,000x^{2}, where x is in meters. Determine the work done by the gas on the bullet as the bullet travels the length of the barrel.

Explanation:

We will calculate the work done as follows.

     W = \int_{0}^{0.54} F dx

         = \int_{0}^{0.54} (14,000 + 10,000x - 26,000x^{2}) dx

         = [14000x + 5000x^{2} - 8666.7x^{3}]^{0.54}_{0}

         = 7560 + 1458 - 1364.69

         = 7653.31 J

or,      = 7.65 kJ       (as 1 kJ = 1000 J)

Thus, we can conclude that the work done by the gas on the bullet as the bullet travels the length of the barrel is 7.65 kJ.

5 0
3 years ago
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