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3 years ago

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Two long, parallel wires carry currents of different magnitudes. If the amount of current in one of the wires is doubled, what h

appens to the magnitude of the force that each wire exerts on the other

1 answer:

Ulleksa [173]3 years ago

3 0

Answer:

The magnitude of the force that each wire exerts on the other will increase by a factor of two.

Explanation:

force on parallel current carrying wire, F = BILsinθ

where;

B is the strength of the magnetic field

L is the length of the wire

I is the magnitude of current on the wire

θ is the angle of inclination of the wire

Assuming B, L and θ is constant, then F ∝ I

F = kI

$\frac{F_1}{I_1} = \frac{F_2}{I_2}$

When the amount of current is doubled in one of the wires, lets say the second wire;

$\frac{F_1}{I_1} = \frac{F_2}{2I_1} \\\\F_2 = \frac{2F_1I_1}{I_1} \\\\F_2 =2F_1$

Also, if will double the amount of current on the first wire, then

F₁ = 2F₂

Therefore, the magnitude of the force that each wire exerts on the other will increase by a factor of two.

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A certain electromagnetic wave traveling in seawater was observed to have an amplitude of 98.02 (V/m) at a depth of 10 m, and an

maksim [4K]

Answer:

The value is $\alpha = 0.002 Np/m$

Explanation:

From the question we are told that

The first amplitude of the wave is $E_{max}1 = 98.02 \ V/m$

The first depth is $D_1 = 10 \ m$

The second amplitude is $E_{max}2 = 81.87 \ (V/m)$

The second depth is $D_2 = 100 \ m$

Generally from the spatial wave equation we have

$v(x) = Ae^{-\alpha d}cos(\beta x + \phi_o)$

=> $\frac{v(x)}{v(x)} =\frac{ Ae^{-\alpha d}cos(\beta x + \phi_o)}{ Ae^{-\alpha d}cos(\beta x + \phi_o)}$

So considering the ratio of the equation for the two depth

$\frac{A}{A_S} = \frac{e^{-D_1 \alpha }}{e^{-D_2 \alpha }}$

=> $\frac{98.02}{81.87} = \frac{e^{-10 \alpha }}{e^{-100 \alpha }}$

=> $\alpha = \frac{0.18}{90}$

=> $\alpha = 0.002 Np/m$

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3 years ago

Scientists are making plans to put a probe in orbit around Earth. They want the probe to enter the orbit shown below.

iris [78.8K]

An arrow which shows the direction that the probe should be moving in order for it to enter the orbit is X.

<h3>What is an orbit?</h3>

An orbit can be defined as the curved path through which a astronomical (celestial) object such as planet Earth, in space move around a Moon, Sun, planet or star.

In this scenario, if the scientists want the probe to enter the orbit they should ensure that probe moves in direction X. This ultimately implies that, the probe must move in the same direction as the orbit, in order to enter it.

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2 years ago

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Determine the index of refraction of glass that is struck by unpolarized light at 53.8 degrees and resulting in a fully polarize

nignag [31]

Answer:

The refractive index of glass, $\mu_{g} = 1.367$

Solution:

Brewster angle is the special case of incident angle that causes the reflected and refracted rays to be perpendicular to each other or that angle of incident which causes the complete polarization of the reflected ray.

To determine the refractive index of glass:

$tan\theta_{P} = \frac{\mu_{g}}{\mu_{a}}$ (1)

where

$\mu_{a}$ = refractive index of glass

$\mu_{g}$ = refractive index of glass

Now, using eqn (1)

$tan{53.8^{\circ}} = \frac{\mu_{g}}{1}$

$\mu_{g} = tan53.8^{\circ}$

$\mu_{g} = 1.367$

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3 years ago

A high diver of mass 51.7 kg steps off a board 10.0 m above the water and falls vertical to the water, starting from rest. If he

zmey [24]

Answer:

851.33 N

Explanation:

Using newton;s equation of motion,

v² = u² + 2gh ......... Equation 1

Where v = final velocity, u = initial velocity, g = acceleration due to gravity, h = height

Given: u = 0 m/s(from rest), h = 10 m g = 9.8 m/s².

Substitute into equation 1

v² = 0² + 2(9.8)(10)

v² = 196

v = √196

v = 14 m/s.

Note: As the Diver touches the water, u = 14 m/s and v = 0 m/s( stopped)

Using,

d = (v+u)t/2 .............. equation 2

Where d = distance moved by the diver in water before its motion stopped, t = time taken before it comes to rest

Given: v = 0 m/s, u = 14 m/s t = 2.10 s

Substitute into equation 2

d = (0+14)2.1/2

d = 14.7 m.

Finally

work done by the water to stop the diver = potential energy of the diver

F×d = mgh'................Equation 3

Where F = force of the diver in water, d = distance of the diver in the water, m = mass of the diver, g = acceleration due to gravity, h' = height of the diver from the point of fall to the point where he comes to rest

making F the subject of the equation,

F = mgh/d ............ Equation 4

Given: m = 51.7 kg, h = 10 m, g = 9.8 m/s², d = 14.7 m.

Substitute into equation 4

F = 51.7(10+14.7)(9.8)/14.7

F = 851.33 N

Hence the upward force the water exert on her = 851.33 N

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3 years ago

An observer on the earth sees a spaceship approaching at 0.54c. The ship then launches an exploration vehicle that, according to

AURORKA [14]

Answer:

Explanation:

Expression for relative velocity

= $\frac{v_1+v_2}{1+\frac{v_1v_2}{c^2} }$

= (.54 + .82 )c/ $1+ \frac{.54 \times.82}{1}$

= 1.36 c / 1.4428

= .94 c

β = .94

6 0

3 years ago