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olganol [36]
3 years ago
5

In a lightning bolt, a large amount of charge flows during a time of 1.2 x 10-3 s. Assume that the bolt can be represented as a

long, straight line of current. At a perpendicular distance of 21 m from the bolt, a magnetic field of 8.4 x 10-5 T is measured. How much charged has flowed during the lightning bolt?
Physics
1 answer:
ohaa [14]3 years ago
6 0

Answer: 10.58 C has flowed during the lightning bolt

Explanation:

Given that;

Time of flow t = 1.2 × 10⁻³

perpendicular distance r = 21 m

Magnetic field B = 8.4 x 10⁻⁵ T

Now lets consider the expression for magnetic field;

B = u₀I / 2πr

the current flow is;

I = ( B × 2πr ) / u₀

so we substitute

I = ( (8.4 x 10⁻⁵) × 2 × 3.14 × 21 ) / 4π ×10⁻⁷

=  0.01107792 / 0.000001256

= 8820 A

Hence the charge flows during lightning bolt  will be;

q = It

so we substitute

q = 8820 × 1.2 × 10⁻³

q = 10.58 C

therefore 10.58 C has flowed during the lightning bolt

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The magnitude of the tension in the string marked A is 52.5N

Generally, the equation for is mathematically given as

Let's take θ be an angle at A

So, tanθ = 3/8

Let's take α be an angle at B (Below X)

tanα = 5/4

Let's take β be an angle at C (Below x)

tanβ = 1/6

First we take the Horizontal Components

74.9cos(9.46°) = Acos(20.6°) + Bcos(51.3°)

By solving the equation, we get

A = 78.9 - 0.668B … (1)

Now, we take the vertical components

74.9sin(9.46°) + Asin(20.6°) = Bsin(51.3°)

By solving the equation, we get

40.07 = 1.015B

B = 39.5N

By substituting the value of B in equation (1)

A = 78.9 - 0.6668× 39.5

A = 52.5N

Hence, the magnitude of the tension in the string marked A is 52.5N

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8 0
2 years ago
A temporary license permits the holder to drive for up to _ days while the application is reviewed
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I drop a meter stick , and catch it as it falls. If it fell exactly 22 cm, how much did it take me to catch the meter stick (wha
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D=Vot+1/2at^2

In this case, there is no initial y velocity so the term Vot=0 so d=1/2at^2

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If the airman had a mass of 80 kg, find the magnitude of the air drag acting on him when he reached terminal velocity of 54 m/s
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The magnitude of the air drag is 784 N

Explanation:

An object falling down reaches the terminal velocity when the magnitude of the air drag acting on it becomes equal to the weight of the object. Mathematically, this condition can be written as:

F_D = mg

where

F_D is the magnitude of the air drag

m is the mass of the object

g is the acceleration of gravity

In this problem, we have

m = 80 kg is the mass of the airman

g=9.8 m/s^2 is the acceleration of gravity

Substituting into the formula, we find:

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