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Degger [83]
2 years ago
9

A 0.4 m long solenoid has a total of 356 turns of wire and carries a current of 79 A. What is the magnitude of the magnetic fiel

d at the center of the solenoid?
Physics
1 answer:
Masja [62]2 years ago
5 0

Answer:

B = 0.088 T

Explanation:

Given that,

The length of a solenoid, l = 0.4 m

No. of turns of wire, N = 356

Current, I = 79 A

We need to find the magnitude of the magnetic field at the center of the solenoid. It is given by the formula.

B=\mu_onI\\\\B=\mu_o \dfrac{N}{l}\times I\\\\B=4\pi \times 10^{-7}\times \dfrac{356}{0.4}\times 79\\\\B=0.088\ T

So, the magnitude of the magnetic field at the center of the solenoid is 0.088 T.

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I know they are stable, have no electrical charge, have interactions with electrons.

Explanation:

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3. Suppose you take a pendulum with length L and mass m having a period T to a
natta225 [31]

(C)

Explanation:

t = 2\pi \sqrt{ \frac{l}{g} }

If g is only 1/6 on another planet, then

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=  \sqrt{6} \: (2\pi \sqrt{ \frac{l}{g} } ) = 2.4  \times t(on \: earth)

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2 years ago
You pick up a 10-newton book off the floor and put it on a shelf 2 meters high. How much work did you do?​
Murljashka [212]

Answer:

20 J

Explanation:

Given:

Weight of the book is, W=10\ N

Height or displacement of the book is, d=2\ m

The work done on the book to raise it to a height of 2 m on a shelf is against gravity. The gravitational force acting on the book is equal to its weight. Now, in order to raise it, an equal amount of force must be applied in the opposite direction.

So, the force applied by me should be equal to weight of the body and in the upward direction. The displacement is also in the upward direction.

Now, work done by the applied force is equal to the product of force applied and displacement of book in the direction of the applied force.

Therefore, work done is given as:

Work=W\times d\\Work=10\times 2=20\ J

Therefore, the work done to raise a book to a height 2 m from the floor is 20 J.

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Valency of oxygen is two. why??​
AlexFokin [52]

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A projectile is shot from the edge of a cliff above the ground level with initial velocity of at an angle with the horizontal. (
Xelga [282]

Answer:

t = √2y/g

Explanation:

This is a projectile launch exercise

a) The vertical velocity in the initial instants (v_{oy} = 0) zero, so let's use the equation

     y =v_{oy} t -1/2 g t²

     y= - ½ g t²

     t = √2y/g

b) Let's use this time and the horizontal displacement equation, because the constant horizontal velocity

     x = vox t

     x = v₀ₓ √2y/g

c) Speeds before touching the ground

     vₓ = vox = constant

     v_{y} = v_{oy} - gt

     v_{y} = 0 - g √2y/g

    v_{y}  = - √2gy

    tan θ = Vy / vx

    θ = tan⁻¹ (vy / vx)

    θ = tan⁻¹ (√2gy / vox)

d) The projectile is higher than the cliff because it is a horizontal launch

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3 years ago
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