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iogann1982 [59]

3 years ago

6

An incredible amount of electrical energy passes down the funnel of a large tornado every second. Measurements taken in Oklahoma

at a distance of 9.00 km from a large tornado showed an almost constant magnetic field of 1.50 · 10 -8 T associated with the tornado. What was the average current going down the funnel? (µo = 4π · 10 -7 T·m/A)

1 answer:

Lorico [155]3 years ago

4 0

Answer:

675 A

Explanation:

B = 1.5 x 10^-8 T

r = 9 km = 9000 m

Let the current be i

The formula for the magnetic filed due to a very long straight conductor is given by

$B =\frac{\mu_{0}}{4\pi }\times \frac{2i}{r}$

$1.5\times 10^{-8} =10^{-7}\times \frac{2i}{9000}$

i = 675 A

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As we know that KE and PE is same at a given position

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now it is given that

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now we will have

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Part b)

KE of SHO at x = A/3

we can use the formula

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jok3333 [9.3K]

Answer:

$S_a_v_e_r_a_g_e=48km/h$

Explanation:

Ok, the average speed can be calculate with the next equation:

$S_a_v_e_r_a_g_e=\frac{Total\hspace{3}distance}{Total\hspace{3}time}$ (1)

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$Total\hspace{3}distance=2*d$

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$Total\hspace{3}time=t1+t2$

From basic physics we know:

$t=\frac{d}{S1}$

so:

$t1=\frac{d}{S1}$

$t2=\frac{d}{S2}$

Using the previous information in equation (1)

$S_a_v_e_r_a_g_e=\frac{2*d}{\frac{d}{S1} +\frac{d}{S2} }=\frac{2*d}{\frac{d*S2+d*S1}{S1+S2} }$

Factoring:

$S_a_v_e_r_a_g_e=\frac{2*S1*S2}{S1+S2}$ (2)

Finally, replacing the data in (2)

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