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hjlf
3 years ago
7

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.
What emf is induced in a square 190km on a side by this rate of change of field?
Physics
1 answer:
Natalka [10]3 years ago
6 0

Answer:

EMF = 1684.67 Volts

Explanation:

As we know that EMF is induced in a closed conducting loop if the flux linked with the loop is changing with time

So we can say

EMF = \frac{d\phi}{dt}

now we have

\phi = BA

here since magnetic field is constant so we have

EMF = A\frac{dB}{dt}

now we have

A = (190 \times 10^3)(190 \times 10^3)

A = 3.61 \times 10^{10} m^2

now we have

EMF = 3.61\times 10^{10} (\frac{2.8 \times 10^{-6}T}{60 s})

EMF = 1684.67 Volts

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An electron moving with a velocity = 5.0 × 107 m/s enters a region of space where perpendicular electric and a magnetic fields a
Cloud [144]

Answer:

magnetic field will allow the electron to go through 2 x 10^{4}  T k

Explanation:

Given data in question

velocity = 5.0 × 10^{7}

electric filed = 10^{4}

To find out

what magnetic field will allow the electron to go through, undeflected

solution

we know if electron move without deflection i.e. net force is zero on electron and we can say both electric and magnetic force equal in magnitude and opposite in directions

so we can also say

F(net) = Fe + Fb i.e. = 0

q V B +  q E = 0

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3 years ago
Kyle, a 95.0 kg football player, leaps straight up into the air (with no horizontal velocity) to catch a pass. He catches the 0.
babymother [125]

Answer:

The initial speed of the ball was 26.2 m/s

Explanation:

When the football player is in the air at his maximum height the vertical component of velocity is zero, To obtain the horizontal velocity when the player catches the ball we need to apply the linear momentum conservation theorem:

m_1*v_{o1}+m_2*v_{o2}=m_t*v_f\\0.430kg*(v)+m_2*(0)=mt*vt

we need to obtain the time taken to go down.

y=Y_o+v_o*t-\frac{1}{2}g*t^2\\\\0=0.589-4.9t^2\\solving:\\t_1=0.346s\\

We have a horizontal displacement and the time taken to stop, so:

v_f=\frac{d}{t}=\frac{0.0409m}{0.346s}=0.118m/s

so:

0.430kg*(v)+m_2*(0)=(m1+m2)*vt\\v=\frac{(95.0kg+0.430kg)*0.118m/s}{0.430kg}\\\\v=26.2m/s

8 0
3 years ago
Approximately what is the smallest detail observable with a microscope that uses ultraviolet light of frequency 1. 72 x 1015 hz?
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The wavelength, which represents the size of the smallest detectable detail that uses ultraviolet light  , is calculated as follows: 3×10^{8} / 1.72×10^{15} or approximately 1.74×10^{-7}m.

The distance between the two positive, two negative, or two minimal points on the waveform is known as the wavelength of the wave. The following formula expresses the relationship between the frequency and wavelength of light:

f = c / λ

where, f = frequency of light

            c = speed of light

            λ = wavelength of light

Given data = f = 1.72×10^{15}Hz

Therefore, λ = 3×10^{8} / 1.72×10^{15}

                  λ = 1.74×10^{-7}m

The wavelength, which represents the size of the smallest detectable detail that uses ultraviolet light  , is calculated as follows: 3×10^{8} / 1.72×10^{15} or approximately 1.74×10^{-7}m.

Learn more about light here;

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Answer:

A) was reusable

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Check this website out for more information about the space shuttle: https://www.nasa.gov/audience/forstudents/k-4/stories/nasa-knows/what-is-the-space-shuttle-k4.html

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