Using the velocity-time graph, the displacement can be calculated by the area under the velocity-time graph. At 3 seconds the total displacement is then equal to (4)(2) + (4 + 2)*1/2 = 11 m. Assuming that the starting point is at x = 0, then the particle at t=3s is at x=11 m.
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Answer:
The capacitance is cut in half.
Explanation:
The capacitance of a plate capacitor is directly proportional to the area A of the plates and inversely proportional to the distance between the plates d. So if the distance was doubled we should expect that the capacitance would be cut in half. That can be verified by the following equation that is used to compute the capacitance in such cases:
C = (\epsilon)*(A/d)
Where \epsilon is a constant that represents the characteristics for the insulator between the plates. A is the area of the plates and d is the distance between them. When we double d we have a new capacitance, given by:
C_new = (\epsilon)*(A/2d)
C_new = (1/2)*[(\epsilon)*(A/d)]
Since C = (\epsilon)*(A/d)] we have:
C_new = (1/2)*C
"Electrostatic forces are attractive or repulsive forces between particles that are caused by their electric charges."
I say around 40% - 60%
https://www.dmv.ca.gov/portal/dmv/detail/teenweb/more_btn6/traffic/traffic
http://www.teendriversource.org/stats/support_teens/detail/57
http://www.rmiia.org/auto/teens/Teen_Driving_Statistics.asp
(I just corrected the question. Sorry if it is still incorrect.)
Hi
The answer to this question is B. Reaction
