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hjlf
3 years ago
10

Currents during lightning strikes can be up to 50000 A (or more!). We can model such a strike as a 49500 A vertical current perp

endicular to the earth's magnetic field, which is about 12 gauss.
a. What is the force on each meter of this current due to the earth's magnetic field?
Physics
1 answer:
Tema [17]3 years ago
3 0

Answer:

59.4 N

Explanation:

The force exerted on a current-carrying wire due to a magnetic field perpendicular to the wire is given by

F=ILB

where

I is the current in the wire

L is the length of the wire

B is the strength of the magnetic field

Here in this problem, we model the strike as a current-carrying wire, so we have:

I = 49,500 A is the current

L = 1 m is the length (we want to find the force per each meter of length)

B=12 G = 12\cdot 10^{-4} T is the strength of the magnetic field

Therefore, the force on each meter of the current due to the magnetic field is:

F=(49,500)(1)(12\cdot 10^{-4})=59.4 N

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An airplane starts at a stand-still and speeds up with an acceleration of 9.0 m/s2. If the runway is 102 m long, how long will i
Keith_Richards [23]

The plane takes 4.76 s to reach the end of the runway

Explanation:

The motion of the airplane is a uniformly accelerated motion (=at constant acceleration), therefore we can solve the problem by using the following suvat equation:

s=ut+\frac{1}{2}at^2

where

s is the distance covered

u is the initial velocity

t is the time elapsed

a is the acceleration

In this problem, we have:

s = 102 m is the lenght of the runway

u = 0 is the initial velocity of the airplane

a=9.0 m/s^2 is its acceleration

Solving for t, we find the time needed to reach the end of the runway:

t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(102)}{9.0}}=4.76 s

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

7 0
3 years ago
Suppose that instead of dropping the rock you throw it downwards so that its speed after falling 7 meters is 23.43 m/s. How much
Shkiper50 [21]

Answer:

3,298.1 Hz

Explanation:

Solution is attached

5 0
4 years ago
Read 2 more answers
By what factor is the intensity of sound at a rock concert louder than that of a whisper when the two intensity levels are 120 d
lyudmila [28]

Answer:

d) 7.94\times 10^{9}

Explanation:

β₁ = sound level of sound at rock concert = 120 dB

β₂ = sound level of sound due to whisper = 21 dB

I₁ = Intensity of sound at rock concert

I₂ = Intensity of sound due to whisper

sound level of sound at rock concert is given as

\beta _{1} = 10 log\left ( \frac{I_{1}}{10^{-12}} \right )

120 = 10 log\left ( \frac{I_{1}}{10^{-12}} \right )

12 = log\left ( \frac{I_{1}}{10^{-12}} \right )               Eq-1

sound level due to whisper is given as

\beta _{2} = 10 log\left ( \frac{I_{2}}{10^{-12}} \right )

21 = 10 log\left ( \frac{I_{2}}{10^{-12}} \right )

2.1 = log\left ( \frac{I_{2}}{10^{-12}} \right )                          Eq-2

subtracting Eq-2 from Eq-1

12 - 2.1 = log\left ( \frac{I_{1}}{10^{-12}} \right ) - log\left ( \frac{I_{2}}{10^{-12}} \right )

9.9 = log\left ( \frac{I_{1}}{I_{2}} \right )

\left ( \frac{I_{1}}{I_{2}} \right ) = 7.94\times 10^{9}

6 0
3 years ago
WILL GIVE BRAINLIEST PLEASE ANSWER ALL QUESTIONS
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As for explaining the first model, there should be the same amount of + and - charges on the peanuts as well as the cat to show that it is balanced and the peanuts are not attracted or repelled by the cat.
5 0
2 years ago
Astronomers use tiny wobbles of distant stars to detect the presence of planets orbiting those stars. This occurs because the st
Zielflug [23.3K]
Orbits or revolves around the sun
5 0
3 years ago
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