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2 years ago

What is the relationship between electric field lines and equipotential surfaces?

1 answer:

5 0

An equipotential surface is a three-dimensional version of equipotential lines. * *Equipotential lines are always perpendicular to electric field lines* . The process by which a conductor can be fixed at zero volts by connecting it to the earth with a good conductor is called grounding.

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A test car travels in a straight line along the x-axis. The graph in the figure shows the car’s position x as a function of time

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If it is s-t graph , point is c
if it is v-t graph , point is e

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2 years ago

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Certain types of atoms break down at a known rate. Scientists can measure the amounts of the broken down products present in a s

sattari [20]

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2 years ago

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A constant force of 4.0 N acts on a variety of objects. The greatest acceleration will occur in the object that has

aleksandr82 [10.1K]

I think it will be the one with the more inertia

im not sure but hope this somewhat helped

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2 years ago

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An object is located 25.0 cm from a convex mirror. The image distance is -50.0 cm. What is the magnification?

Lynna [10]

Answer:

$\boxed{\sf Magnification \ (m) = 2}$

Given:

Object distance (u) = 25.0 cm

Image distance (v) = -50.0 cm

To Find:

Magnification (m)

Explanation:

$\boxed{\bold{\sf Magnification \: (m) = - \frac{Image \: distance \: (v)}{Object \: distance \: (u)}}}$

Substituting values of Image distance(v) & Object distance (u) in the equation:

$\sf \implies m = - \frac{( - 50)}{25}$

-(-50) = 50:

$\sf \implies m = \frac{50}{25}$

$\sf \implies m = \frac{2 \times \cancel{25}}{ \cancel{25}}$

$\sf \implies m = 2$

4 0

3 years ago

Determine the instantaneous velocity of the car at t = 4.7 s, using time intervals of 0.40 s, 0.20 s, and 0.10 s. (In order to b

weeeeeb [17]

Answer:

4.408 m/s, 4.102 m/s, 4.026 m/s

Explanation:

The question is incomplete. The text of the original question states:

A race car moves such that its position fits the relationship

$x=(4.0 m/s)t + (0.85 m/s^3) t^3$

where x is measured in meters and t in seconds. Determine the instantaneous velocity of the car at t = 4.7 s, using time intervals of 0.40 s, 0.20 s, and 0.10 s.

We can find the instantanoues velocity of the car at any time t by calculating the derivative of the position, so we find:

$v(t) = x'(t) = 4.0 m/s + 3\cdot (0.85 m/s^2) t^2 = 4.0 m/s + (2.55 m/s^2) t^2$

And now we just need to substitute t=0.40 s, 0.20 s, and 0.10 s to find the corresponding instantaneous velocity:

$v(0.40) = 4.0 + 2.55 (0.40)^2 = 4.408 m/s\\v(0.20) = 4.0 + 2.55 (0.20)^2 = 4.102 m/s\\v(0.10) = 4.0 + 2.55 (0.10)^2 = 4.026 m/s$

5 0

3 years ago