Ask question

Ask question

All categories

3 years ago

6

The magnetic field at the earth's surface can vary in response to solar activity. During one intense solar storm, the vertical c

omponent of the magnetic field changed by 2.8 μT per minute, causing voltage spikes in large loops of the power grid that knocked out power in parts of Canada. You may want to review (Pages 815 - 817) . Part A What emf is induced in a square 110 km on a side by this rate of change of field

1 answer:

Flauer [41]3 years ago

5 0

Answer:

$EMF = 33880 Volts$

Explanation:

As per Faraday's law of Electromagnetic induction we know that

Rate of change in magnetic flux will induce EMF in the closed conducting loop

so we have

$\phi = B.A$

now we have

$A = (110 \times 10^3)(110 \times 10^3)$

$A = 1.21 \times 10^{10}$

now we have

$\phi = B(1.21 \times 10^{10})$

now the induced EMF through this loop is given as

$EMF = (\frac{dB}{dt})(1.21 \times 10^{10})$

$EMF = (2.8 \times 10^{-6})(1.21 \times 10^{10})$

$EMF = 33880 Volts$

You might be interested in

A set of statements that belong together as a group and contribute to the function definition is known as a(n) ________.

dsp73

Block. hope this helps.

5 0

3 years ago

When the tube is filled with mercury vapor, as in this case, a sharp drop in the collected current is observed when the accelera

Nata [24]

Answer:

The energy absorbed by the atomic electrons in the mercury atom is $7.84 \times 10^{-19}$ J

Explanation:

Given:

Potential $V = 4.9$ V

According to the conservation law,

Loss in kinetic energy = Gain in potential energy

Here, energy absorbed by the atomic electrons is given by,

$E = eV$

Where $e = 1.6 \times 10^{-19} C$ ( charge of electron )

$E = 1.6 \times 10^{-19 } \times 4.9$

$E = 7.84 \times 10^{-19}$ J

Therefore, the energy absorbed by the atomic electrons in the mercury atom is $7.84 \times 10^{-19}$ J

5 0

3 years ago

A pitcher delivers a fast ball with a velocity of 43 m/s to the south. the batter hits the ball and gives it a velocity of 51 m/

hoa [83]

Assuming north as positive direction, the initial and final velocities of the ball are:
$v_i=-43 m/s$ (with negative sign since it is due south)
$v_f=+51 m/s$
the time taken is $t=1.0 ms=0.001 s$ , so the average acceleration of the ball is given by
$a= \frac{v_f-v_i}{t}= \frac{51 m/s-(-43 m/s)}{0.001 s}=9.4 \cdot 10^4 m/s^2$
And the positive sign tells us the direction of the acceleration is north.

4 0

3 years ago

The density of a sample can be obtained by dividing its __________ by its _______________.

mixer [17]

<span>Density can be calculated and found by dividing the sample's mass by its volume. D=m/v</span>

3 0

3 years ago

A planet is 10 light years away from Earth. What speed would you need to go for a trip to the planet and back to take only 5 yea

viva [34]

Answer:

a. speed, v = 0.97 c

b. time, t' = 20.56 years

Given:

t' = 5 years

distance of the planet from the earth, d = 10 light years = 10 c

Solution:

(a) Distance travelled in a round trip, d' = 2d = 20 c = L'

Now, using Length contraction formula of relativity theory:

$L'' = L'\sqrt{1 - \frac{v^{2}}{c^{2}}}$ (1)

time taken = 5 years

We know that :

time = $\frac{distance}{speed}$

5 = $\frac{L''}{v}$ (2)

Dividing eqn (1) by v on both the sides and substituting eqn (2) in eqn (1):

$\frac{L'\sqrt{1 - \frac{v^{2}}{c^{2}}}}{v} = 5$

$\frac{20'\sqrt{1 - \frac{v^{2}}{c^{2}}}}{v} = 5$

Squaring both the sides and Solving above eqution, we get:

v = 0.97 c

(b) Time observed from Earth:

Using time dilation:

$t'' = \frac{t'}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}$

$t'' = \frac{5}{\sqrt{1 - \frac{(0.97c)^{2}}{c^{2}}}}$

Solving the above eqn:

t'' = 20.56 years

4 0

3 years ago