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Delicious77 [7]
4 years ago
12

A 10 m long high tension power line carries a current of 20 A perpendicular to Earth's magnetic field of 5.5 x10⁻⁵ T. What is th

e magnetic force experienced by the power line? ______ N *
Physics
1 answer:
den301095 [7]4 years ago
7 0

Answer:

Explanation:

Given that,

The length of tension wire is 10m

L = 10m

It caries a current of 10A

I = 10A

The current is perpendicular to the magnetic field

B = 5.5 × x10⁻⁵ T.

Then, θ = 90°

The magnetic force that act on the line is?

F = i(L × B)

L × B = LB Sinθ

Where

I is the current

L is the length of the wire

B is the magnetic field

θ is the angle between the field and the direction of the length.

Then,

F = I × L × B × Sinθ

F = 10 × 10 × 5.5 × x10⁻⁵ × Sin 90

F = 5.5 × x10⁻³ N

Then, the magnetic force experience on the line is 5.5 × x10⁻³N

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A block of mass m=16.8 kg is sliding on a surface with initial velocity v=23.2 m/s. The block has a coefficient of kinetic frict
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Answer:

t = 23.92 s

Explanation:

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  and the y-axis in the direction perpendicular to it.

Forces acting on the block

W: Weight of the block : In vertical direction, downward

FN : Normal force : perpendicular to the floor, upward

fk :  Kinetic friction force: parallel to the floor  and opposite to the movement

F = 86.4 N , in the direction of the motion

Calculated of the W

W= m*g

W=  16.8 kg* 9.8 m/s² = 164.64 N

Calculated of the FN  

We apply the formula (1)  

∑Fy = m*ay ay = 0  

FN - W = 0  

FN = W  

FN =  164.64 N

Calculated of the fk

fk  = μk*FN

fk  = 0.426* 164.64 N

fk  = 70.13 N

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

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

F - fk = m*a

86.4 -70.13  = (16.8)*(-a)

16.26 =  (16.8)*(-a)

a = -(16.26 )/ (16.8)

a = - 0.97 m/s²

Kinematics of the block

Because the block moves with uniformly accelerated movement we apply the following formula :

vf = v₀ + a*t   Formula (2)

Where:  

t: time interval  (m)

v₀: initial speed  (m/s)

vf: final speed   (m/s)

Data:

v₀ = 23.2 m/s

vf = 0

a =  -0.97 m/s²  

Time it takes for the block to stop

We replace data in the formula (2)  to calculate the time

vf= v₀+a*t

0 = 23.2+( -0.97)*t

(0.97)*t  = 23.2

(0.97)*t  = 23.2

t = 23.2 / (0.97)

t = 23.92 s

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