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Kazeer [188]
2 years ago
14

What is the magnitude of the electric field at the dot in the figure?

Physics
1 answer:
Zielflug [23.3K]2 years ago
8 0

The magnitude of the electric field at the dot is :  10⁴ v/m

Given that there are three equipotential lines with equal spacing,we will apply the the relationship between P.D and electric field

<h3>Determine the magnitude of the electric field at the dot </h3>

change in voltage  = E .d

100 - 0 = E * ( 1 * 10⁻² m ) ----- ( 1 )

From equation ( 1 )

The magnitude of E = 100 v / ( 1 * 10⁻² m )

                                 = 10⁴ v/m

Hence we can conclude that The magnitude of the electric field at the dot is :  10⁴ v/m

Learn more about electric field : brainly.com/question/14372859

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A 3.15-kg object is moving in a plane, with its x and y coordinates given by x = 6t2 − 4 and y = 5t3 + 6, where x and y are in m
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a_y = \frac{d}{dt}(\frac{dy}{dt} )\\ \\a_y = \frac{d}{dt}(5t^{3} +6 )\\\\a_y = \frac{d}{dt}(15t^{2}   )\\\\a_y = 30t\\a_y \ at \ t= 2.15s; a_y = 30(2.15)\\a_y = 64.5m/s^2

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R = \sqrt{37.8^2+203.18^2}\\ \\R = \sqrt{1428.84+41,282.11}\\ \\R = \sqrt{42.710.95}\\ \\R = 206.67N

Hence, the magnitude of the net force acting on this object at t = 2.15 s is approximately 206.67N

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