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PIT_PIT [208]
2 years ago
5

During lightning strikes from a cloud to the ground, currents as high as 2.50×10^4 Amps can occur and last for about 40.0 micros

econds . How much charge is transferred from the cloud to the earth during such a strike?
Physics
1 answer:
dangina [55]2 years ago
6 0

Answer:

1 C

Explanation:

The intensity of electric current is defined as

I=\frac{q}{t}

where

I is the current

q is the amount of charge transferred

t is the time interval during which the charge is transferred

For the lightning in this problem, we have

I=2.50\cdot 10^4 A is the current

t=40.0 \mu s = 40.0\cdot 10^{-6} s is the time interval

Solving the formula for q, we find the amount of charge transferred:

q=I t = (2.50\cdot 10^4 A)(40.0\cdot 10^{-6}s)=1 C

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Papessa [141]

The initial speed of the bolt is not 58.86 m/s.  

Let a be the acceleration of the rocket.  

During the 4 sec lift off, the rocket has reached a height of  

h = (1/2)*a*t^2  

with t=4,  

h = (1/2)*a^16  

h = 8*a  

Its velocity at 4 sec is  

v = t*a  

v = 4*a  

The initial velocity of the bolt is thus 4*a.  

During the 6 sec fall, the bolt has the initial velocity V0=-4*a and it drops a total height of h=8*a. From the equation of motion,  

h = (1/2)*g*t^2 + V0*t  

Substituting h0=8*a, t=6 and V0=-4*a into it,  

8*a = (1/2)*g*36 - 4*a*6  

Solving for a  

a = 5.52 m/s^2

6 0
2 years ago
What is the magnitude (in N/C) and direction of an electric field that exerts a 3.50 ✕ 10−5 N upward force on a −1.55 µC charge?
IRISSAK [1]

Answer:

The magnitude of electric field is  22.58 N/C

Solution:

Given:

Force exerted in upward direction, \vec{F_{up}} = 3.50\times 10^{- 5} N

Charge, Q = - 1.55\micro C = - 1.55\times 10^{- 6} C

Now, we know by Coulomb's law,

F_{e} = \frac{1}{4\pi\epsilon_{o}\frac{Qq}{R^{2}}

Also,

Electric field, E = \frac{1}{4\pi\epsilon_{o}\frac{q}{R^{2}}

Thus from these two relations, we can deduce:

F = QE

Therefore, in the question:

\vec{E} = \frac{\vec F_{up}}{Q}

\vec{E} = \frac{3.50\times 10^{- 5}}{- 1.55\times 10^{- 6}}

\vec{E} = - 22.58 N/C

Here, the negative side is indicative of the Electric field acting in the opposite direction, i.e., downward direction.

The magnitude of the electric field is:

|\vec{E}| = 22.58\ N/C

6 0
2 years ago
How much force is needed to accelerate a 1750-kg car at a rate of 3 m/s2?
Rina8888 [55]
The force needed is 5250N
5 0
3 years ago
Newly discovered planet has twice the mass and three times the radius of the earth. What is the free-fall acceleration at its su
skad [1K]

Answer:

g_n=\dfrac{2}{9}g

Explanation:

M = Mass of Earth

G = Gravitational constant

R = Radius of Earth

The acceleration due to gravity on Earth is

g=\dfrac{GM}{R^2}

On new planet

g_n=\dfrac{G2M}{(3R)^2}\\\Rightarrow g_n=\dfrac{2GM}{9R^2}

Dividing the two equations we get

\dfrac{g_n}{g}=\dfrac{\dfrac{2GM}{9R^2}}{\dfrac{GM}{R^2}}\\\Rightarrow \dfrac{g_n}{g}=\dfrac{2}{9}\\\Rightarrow g_n=\dfrac{2}{9}g

The acceleration due to gravity on the other planet is g_n=\dfrac{2}{9}g

4 0
2 years ago
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galben [10]

Answer:

3.51s

Explanation:

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