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rewona [7]
3 years ago
6

During lightning strikes from a cloud to the ground, currents as high as 2.50×104 A can occur and last for about 40.0 μs . How m

uch charge is transferred from the cloud to the earth during such a strike?
Physics
1 answer:
Nezavi [6.7K]3 years ago
3 0

Answer:

<h3>The charge transferred from the cloud to earth is 1 Coulomb.</h3>

Explanation:

Given :

Current I = 2.5 \times 10^{4} A

Time t = 40 \times 10^{-6} sec

We know that the current is the rate of flow of charge.

From the formula of current,

<h3>  I = \frac{Q}{t}</h3>

Where Q = charge transfer between cloud and earth.

 Q =I t

Q = 2.5 \times 10^{4} \times 40 \times 10^{-6}

Q = 1 C

Hence, the charge transferred from the cloud to earth is 1 Coulomb.

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Boiling water

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The 1350-kg car has a velocity of 24 km/h up the 8-percent grade when the driver applies more power for 18 s to bring the car up
Brrunno [24]

Answer:

Explanation:

We shall apply concept of impulse to solve the problem  .

Impulse = force x time

impulse = change in momentum

force x time = change in momentum

initial speed u = 24 km/h = 6.67 m /s

final speed v = 65 km/h = 18.05 m /s

change in momentum = m v - mu

= m ( v-u )

= 1350 ( 18.05 - 6.67 )

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3 years ago
You are running at a speed of 10km/h and hit a patch of mud. Two seconds later you speed is 8km/h. What is your acceleration in
Vlad1618 [11]

Answer:

0.28 m/s^2

Explanation:

Acceleration is given by

a=\frac{v-u}{t}

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u is the initial velocity

v is the final velocity

t is the time interval

In this problem:

u = 10 km/h \cdot \frac{1000 m/km}{3600 s/h}=2.78 m/s is the initial velocity

v = 8 km/h \cdot \frac{1000 m/km}{3600 s/h}= 2.22 m/s is the final velocity

t = 2 s is the time

Substituting, we find the acceleration:

a=\frac{2.78-2.22}{2}=0.28 m/s^2

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Read 2 more answers
The magnitude of the velocity vector of the car is ∣∣v→∣∣ = 78 ft/s. If the vector v→ forms an angle θ = 0.09 rad with the horiz
kotegsom [21]

Answer:

\vec{v} = (77.68~{\rm ft/s})\^i + (7.01~{\rm ft/s})\^j

Explanation:

The x- and y- components of the velocity vector can be written as following:

\vec{v}_x = ||\vec{v}||\cos(\theta)\^i

\vec{v}_y = ||\vec{v}||\sin(\theta)\^j

Since the angle θ and the magnitude of the velocity is given, the vector representation can be written as follows:

\vec{v} = 78\cos(0.09)\^i + 78\sin(0.09)\^j\\\vec{v} = (77.68~{\rm ft/s})\^i + (7.01~{\rm ft/s})\^j

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2 years ago
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