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marissa [1.9K]
2 years ago
8

Assuming a current is flowing, what increases the strength of the magnetic field of a coiled electrical wire? Select all that ap

ply. Straightening the wire changing the direction of the current flow adding more coils to the wire placing a metal bar within the coils stopping the current.
Physics
1 answer:
goblinko [34]2 years ago
8 0

The strength of the magnetic field is increased by adding more coils to the wire and placing a metal bar within the coils.

The given problem is based on the concept and fundamentals of the magnetic strength in a current-carrying coil of wire. The current-carrying wire in a coiled form is known as Solenoid.

And the mathematical expression for the magnetic field in a solenoid is,[B = \dfrac{\mu_{0} \times N \times i}{L}

Here,

B is the strength of the magnetic field.

\mu_{0} is the magnetic permeability of free space.

N is the number of coils.

The strength of the magnetic field will increase by increasing the number of coils and placing the metallic slab (it will increase magnetic permeability) within the coils.

Thus, we can conclude that the strength of the magnetic field is increased by adding more coils to the wire and placing a metal bar within the coils.

learn more about the magnetic field here:

brainly.com/question/19542022

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A 4.25 kg block is sent up a ramp inclined at an angle theta=37.5° from the horizontal. It is given an initial velocity ????0=15
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Answer:

d = 11.79 m

Explanation:

Known data

m=4.25 kg  : mass of the block

θ =37.5°  :angle θ of the ramp with respect to the horizontal direction

μk= 0.460  : coefficient of kinetic friction

g = 9.8 m/s² : acceleration due to gravity

Newton's second law:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration  (m/s²)

We define the x-axis in the direction parallel to the movement of the block on the ramp and the y-axis in the direction perpendicular to it.

Forces acting on the block

W: Weight of the block : In vertical direction

N : Normal force : perpendicular to the ramp

f : Friction force: parallel to the ramp

Calculated of the W

W= m*g

W=  4.25 kg* 9.8 m/s² = 41.65 N

x-y weight components

Wx= Wsin θ= 41.65*sin 37.5° = 25.35 N

Wy= Wcos θ =41.65*cos 37.5° =33.04 N

Calculated of the N

We apply the formula (1)

∑Fy = m*ay    ay = 0

N - Wy = 0

N = Wy

N = 33.04 N

Calculated of the f

f = μk* N= 0.460*33.04

f = 15.2 N

We apply the formula (1) to calculated acceleration of the block:

∑Fx = m*ax  ,  ax= a  : acceleration of the block

-Wx-f = m*a

 -25.35-15.2 = (4.25)*a

-40.55 =  (4.25)*a

a = (-40.55)/ (4.25)

a = -9.54 m/s²

Kinematics of the block

Because the block moves with uniformly accelerated movement we apply the following formula to calculate the final speed of the block :

vf²=v₀²+2*a*d Formula (2)

Where:  

d:displacement  (m)

v₀: initial speed  (m/s)

vf: final speed   (m/s)

Data:

v₀ = 15 m/s

vf = 0

a = -9.54 m/s²

We replace data in the formula (2)  to calculate the distance along the ramp the block reaches before stopping (d)

vf²=v₀²+2*a*d

0 = (15)²+2*(-9.54)*d

2*(9.54)*d =   (15)²

(19.08)*d = 225

d = 225 / (19.08)

d = 11.79 m

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