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patriot [66]
3 years ago
6

7. An automobile with a radio antenna 1.0 m long travels at 100.0 km/h in a location where the Earth’s horizontal magnetic field

is 5.5×10−5T. What is the maximum possible emf induced in the antenna due to this motion?
Physics
1 answer:
VashaNatasha [74]3 years ago
3 0

Answer:

0.00479 volts

Explanation:

From Faraday's law of electromagnetic induction, the induced emf equals the change in magnitude flux (magnetic field strength multiplied by the area = BA) divided by the time change

Therefore we have the equation

EMF = BA÷t

Since area A = 2πr

EMF = Bπr²÷t

B = 5.5×10^(-5)

Velocity = 100km/h = 27.7778m/s

r = 1m, t = r÷V = 0.036

EMF = Bπr²÷t = (5.5×10^(-5) x π x (1)²)÷0.036 = 0.00479 Volts

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Study the velocity vs. time graph shown.
laiz [17]

For velocity vs Time graphs, the displacement of the object from 2 seconds to 6 seconds is 30 m.

<h3>What is displacement?</h3>

The displacement is the shortest distance travelled by the particle. It is the vector quantity which re[presents both the magnitude and direction.

In velocity time graphs, the displacement is the area under the curve of the graph on the x axis.

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A line starts at (0, 2) and ends at (6, 8) in v-t graph

Displacement is equal to the area of a triangle and a rectangle formed under the line.

Area = 1/2 base x height + length x breadth

Area = 1/2 x 6x 6 + 6x2

Area = 18 +12

Area = 30 m

Thus, the displacement is 30 m.

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6 0
2 years ago
Why does the ball orbit the Earth when launched from the theoretical cannon of Newton?
Lyrx [107]

The ball orbit the Earth, when launched from the theoretical cannon of Newton, is option B. it is magnetically attracted.

<h3>Newton's Cannonball:</h3>

Newton's cannonball was a hypothetical situation. Isaac Newton once proposed that gravity, which he believed to be a universal force, was the primary factor behind the planetary motion. In this experiment, Newton imagines projecting a stone or a cannonball onto the summit of a very tall mountain. The body should move away from Earth in the direction it was projected if there were no effects from gravity or air resistance.

Depending on the projectile's initial velocity and the gravitational force acting on it, the bullet will travel in a different direction. Low speeds result in a simple fallback to Earth. The Earth's surface causes the cannonball to deviate from its elliptical route.

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4 0
2 years ago
The drawings show two examples in which a ray of light is refracted at the interface between two liquids. In each example the in
Nataly [62]

Answer:

Possible options:

A. nA, nC, nB

B. nA, nB, nC

C. nC, nB, nA

D. nC, nA, nB

E. nB, nA, nC

Answer is D

Explanation:

When the light is refracted into liquid B it is bent away from the normal, so that nA > nB. When the light is refracted into liquid C it is bent toward the normal, so that nC > nA. Therefore, we conclude that nC > nA > nB

6 0
3 years ago
Help me guys<br><br>Class IX​
goldfiish [28.3K]

Answer:

with right hand grip rule

3. A- south

B- north

C- north

D- south

E- south

F- north

sorry idk what 1st & 2nd question means

4 0
3 years ago
A coil is wrapped with 300 turns of wire on the perimeter of a circular frame (radius = 8.0 cm). Each turn has the same area, eq
MAVERICK [17]

Answer:

Approximately 18 volts when the magnetic field strength increases from \rm 20\; mT to \rm 80\;mT at a constant rate.

Explanation:

By the Faraday's Law of Induction, the EMF \epsilon that a changing magnetic flux induces in a coil is:

\displaystyle \epsilon = N \cdot \frac{d\phi}{dt},

where

  • N is the number of turns in the coil, and
  • \displaystyle \frac{d\phi}{dt} is the rate of change in magnetic flux through this coil.

However, for a coil the magnetic flux \phi is equal to

\phi = B \cdot A\cdot \cos{\theta},

where

  • B is the magnetic field strength at the coil, and
  • A\cdot \cos{\theta} is the area of the coil perpendicular to the magnetic field.

For this coil, the magnetic field is perpendicular to coil, so \theta = 0 and A\cdot \cos{\theta} = A. The area of this circular coil is equal to \pi\cdot r^{2} = \pi\times 8.0\times 10^{-2}\approx \rm 0.0201062\; m^{2}.

A\cdot \cos{\theta} = A doesn't change, so the rate of change in the magnetic flux \phi through the coil depends only on the rate of change in the magnetic field strength B. The size of the magnetic field at the instant that B = \rm 50\; mT will not matter as long as the rate of change in B is constant.

\displaystyle \begin{aligned} \frac{d\phi}{dt} &= \frac{\Delta B}{\Delta t}\times A \\&= \rm \frac{80\times 10^{-3}\; T- 20\times 10^{-3}\; T}{20\times 10^{-3}\; s}\times 0.0201062\;m^{2}\\&= \rm 0.0603186\; T\cdot m^{2}\cdot s^{-1}\end{aligned}.

As a result,

\displaystyle \epsilon = N \cdot \frac{d\phi}{dt} = \rm 300 \times 0.0603186\; T\cdot m^{2}\cdot s^{-1} \approx 18\; V.

9 0
3 years ago
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