Assume Earth and a cloud layer 0.900 Km above the Earth can be treated as plates of a parallel-plate capacitor. a. If the cloud layer has an area of 2.00 × 104 m2, what is the capacitance?
b. If an electric field strength of 2.0 × 105 N/C causes the air to conduct charge (lightning), what charge can the cloud hold?
c. Describe what must happen to its molecules for air to conduct electricity.
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(A)-1.96e-10 nanoFarads (B)- 0.03528 C (C)- The flow of electricity is the flow of electrons through a medium. The air has to partially ionize in order to allow for the flow of electrons.Explanation:

(A) Capacitance can be calculated using or . This results in a value of 1.96e-12 nanofarads

(B) The total charge can be calculated using the equation C*V=Q

We have to find the voltage across the capacitor. To do this use the equation E*D=V. This results in a value of 1.8e8 Volts. Plug this value and the capacitance value into the equation C*V=Q. This will result in a value of 0.03528 coulombs of charge.

(C) The flow of electricity is the flow of electrons through a medium. The air has to partially ionize in order to allow for the flow of electrons.

weather forecasts are done by collecting data about the current state of the atmosphere mainly the temperature, humidity and wind and using meteorology, to determine the atmosphere that evolves over time. the h and l mean the high and low pressure. the red strip is a warm front. the blue strip is a cold front. the blue and red strip is a stationary front. the purple strip is a occluded front.

explanation:

i believe the answer is a