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Delicious77 [7]
2 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]2 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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When a vertical beam of light passes through a transparent medium, the rate at which its intensity I decreases is proportional t
antoniya [11.8K]

Answer:

Intensity of beam 18 feet below the surface is about 0.02%

Explanation:

Using Lambert's law

Let dI / dt = kI, where k is a proportionality constant, I is intensity of incident light and t is thickness of the medium

then dI / I = kdt

taking log,

ln(I) = kt + ln C

I = Ce^kt

t=0=>I=I(0)=>C=I(0)

I = I(0)e^kt

t=3 & I=0.25I(0)=>0.25=e^3k

k = ln(0.25)/3

k = -1.386/3

k = -0.4621

I = I(0)e^(-0.4621t)

I(18) = I(0)e^(-0.4621*18)

I(18) = 0.00024413I(0)

Intensity of beam 18 feet below the surface is about 0.2%

3 0
2 years ago
A semi is traveling down the highway at a velocity of v = 26 m/s. The driver observes a wreck ahead, locks his brakes, and begin
Dovator [93]

Answer:

fcosθ + Fbcosθ  =Wtanθ

Explanation:

Consider the diagram shown in attachment

fx= fcosθ (fx: component of friction force in x-direction ; f: frictional force)

Fbx= Fbcosθ ( Fbx: component of braking force in x-direction ; Fb: braking force)

Wx= Wtanθ (Wx: component of weight in x-direction ; W: Weight of semi)

sum of x-direction forces = 0

fx+ Fbx=Wx

fcosθ + Fbcosθ  =Wtanθ

7 0
2 years ago
Wrapping paper is being unwrapped from a 5.0-cm radius tube, free to rotate on its axis. if it is pulled at the constant rate of
lisov135 [29]
So the equation for angular velocity is

Omega = 2(3.14)/T

Where T is the total period in which the cylinder completes one revolution.

In order to find T, the tangential velocity is

V = 2(3.14)r/T

When calculated, I got V = 3.14

When you enter that into the angular velocity equation, you should get 2m/s
5 0
2 years ago
A windowpane is half a centimeter thick and has an area of 1.0 m2. The temperature difference between the inside and outside sur
polet [3.4K]

To solve this problem it is necessary to apply the concepts related to the heat flux rate expressed in energetic terms. The rate of heat flow is the amount of heat that is transferred per unit of time in some material. Mathematically it can be expressed as:

\frac{Q}{t} = \frac{kA}{L} (T_H - T_C)

Where

k = 0.84 J/s⋅m⋅°C (The thermal conductivity of the material)

A = 1m^2 Area

L = 5*10^{-3}m Length

T_H= Temperature of the "hot"reservoir

T_C= Temperature of the "cold"reservoir

Replacing with our values we have that,

\frac{Q}{t} = \frac{kA}{L} (T_H - T_C)

\frac{Q}{t} = \frac{(0.84)(1)}{0.005} (15)

\frac{Q}{t} = 2520J/s

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3 0
2 years ago
An athlete prepares to throw a 2.0-kilogram discus. His arm is 0.75 meters long. He spins around several times with the discus a
wlad13 [49]
Centripetal Force (Fcp) = ?

His arm length = Radius (R) = 0.75 m

Discus velocity = Linear Velocity (V) = 5 m/s

Discus mass (m) = 2 kg

Centripetal Acceleration (Acp) = V^2/R or W^2 x R
In this case i will use the V^2/R formula, because it uses the discus velocity (V).

fcp = m \times acp \\ fcp = m \times {v}^{2} \div r \\ fcp = 2 \times {5}^{2} \div 0.75 \\ fcp = 2 \times 25 \div 0.75 \\ fcp = 50 \div 0.75
fcp = 66.666... = 66 \: newtons

Answer: Last option, 66 N.
7 0
2 years ago
Read 2 more answers
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