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Dvinal [7]
3 years ago
6

Which statement best explains the movement of electric current from the clouds to the ground during a lightning storm? .

Physics
2 answers:
SVEN [57.7K]3 years ago
5 0

Answer:

Correct Answer is option C

Explanation:

Option A cannot be correct because if both earth and clouds are negatively charged then they will repel each other and electron cannot flow through it because electron is negatively charged and it will get attracted towards positively charged terminal. Hence electric current will not flow between them.

Option B

cannot be correct because if both earth and clouds are positively charged then they will repel each other and electron cannot flow through it.Hence electric current will not flow between them.

Option C

It is correct option because flow of current means flow of electron. so electron which are negatively charged particle will move from cloud to positively charged ground.

Option D

If clouds are neutral then electron will not be there and hence current will not flow.

olga2289 [7]3 years ago
3 0
"The <span>ground is positively charged and the clouds are negatively charged " is the statement among the statements given in the question that </span><span>best explains the movement of electric current from the clouds to the ground during a lightning storm. The correct option among all the options that are given in the question is the third option or option "C". </span>
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Three particles lie in the xy plane. Particle 1 has mass m1 = 6.7 kg and lies on the x-axis at x1 = 4.2 m, y1 = 0. Particle 2 ha
krek1111 [17]

Answer:

F=18.58\times 10^{-11}\ N

\theta=30.276^{\circ}

Explanation:

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mass of first particle, m_1=6.7\ kg

mass of second particle, m_2=5.1\ kg

mass of third particle, m_3=3.7\ kg

coordinate position of first particle in meters, (x_1,y_1)\equiv(4.2,0)

coordinate position of second particle in meters, (x_2,y_2)\equiv(0,2.8)

coordinate position of third particle in meters, (x_3,y_3)\equiv(0,0)

<u>Now, gravitational force on particle 3 due to particle 1:</u>

F_{31}=G\frac{m_1.m_3}{r_{31}^2}

F_{31}=6.67\times 10^{-11} \times \frac{6.7\times 3.7}{4.2^2}

F_{31}=9.37\times 10^{-11}\ N

towards positive Y axis.

<u>gravitational force on particle 3 due to particle 2:</u>

F_{32}=G\frac{m_2.m_3}{r_{21}^2}

F_{32}=6.67\times 10^{-11} \times \frac{5.1\times 3.7}{2.8^2}

F_{32}=16.05\times 10^{-11}\ N

towards positive X axis.

<u>Now the net force</u>

F=\sqrt{F_{31}\ ^2+F_{32}\ ^2}

F=\sqrt{(10^{-11})^2(9.37^2+16.05^2)}

F=18.58\times 10^{-11}\ N

<em>For angle in counterclockwise direction from the +x-axis</em>

tan\theta=\frac{9.37\times 10^{-11}}{16.05\times 10^{-11}}

\theta=30.276^{\circ}

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