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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)
Physics
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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What is the smallest part of a chemical compound that has all the properties
mel-nik [20]

Answer:

D. A Molecule

Explanation:

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5 0
2 years ago
A helicopter starting on the ground is rising directly into the air at a rate of 25 ft/s. You are running on the ground starting
rusak2 [61]

Answer:

The rate of change of the distance between the helicopter and yourself (in ft/s) after 5 s is \sqrt{725} ft/ sec

Explanation:

Given:

h(t) =  25 ft/sec

x(t) = 10 ft/ sec

h(5) = 25 ft/sec . 5 = 125 ft

x(5) = 10 ft/sec . 5 = 50 ft

Now we can calculate the distance between the person and the helicopter by using the Pythagorean theorem

D(t) = \sqrt{h^2 + x^2}

Lets find the derivative of distance with respect to time

\frac{dD}{dt} (t)  = \frac{2h \cdot \frac{dh}{dt} +2x \cdot\frac{dx}{dt}} {2\sqrt{h^2 + x^2}}

Substituting the values of h(t) and  x(t) and simplifying we get,

\frac{dD}{dt}(t) = \frac{50t \cdot \frac{dh}{dt} + 20 \cdot \frac{dx}dt}{2\sqrt{625\cdot t^2 + 100 \cdot t^2}}

\frac{dh}{dt} = 25ft/sec

\frac{dx}{dt} = 10 ft/sec

\frac{Dd}{dt} (t) = \frac{1250t +200t}{2\sqrt{725}t}  = \frac{725}{\sqrt{725}}  = \sqrt{725} ft / sec

5 0
3 years ago
A high jumper jumps over a bar that is 2 m above the mat. With what velocity does the jumper strike the mat in the landing area?
docker41 [41]

Answer:

The velocity with which the jumper strike the mat in the landing area is 6.26 m/s.

Explanation:

It is given that,

A high jumper jumps over a bar that is 2 m above the mat, h = 2 m

We need to find the velocity with which the jumper strike the mat in the landing area. It is a case of conservation of energy. let v is the velocity. it is given by :

v=\sqrt{2gh}

g is acceleration due to gravity

v=\sqrt{2\times 9.81\ m/s^2\times 2\ m}

v = 6.26 m/s

So, the velocity with which the jumper strike the mat in the landing area is 6.26 m/s. Hence, this is the required solution.

8 0
3 years ago
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