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Sergeu [11.5K]
2 years ago
9

Points a and b lie in a region where the y-component of the electric field is Ey=α+β/y2. The constants in this expression have t

he values α = 600 N/C and β = 5.00 N⋅m2/C. Points a and b are on the y-axis. .
Point a is at y = 2.00 cm and point b is at y = 3.00 cm. What is the potential difference Va−Vb between these two points.?
Physics
1 answer:
Drupady [299]2 years ago
3 0

Answer:

V_{a} - V_{b} = 89.3

Explanation:

The electric potential is defined by

         V_{b} - V_{a} = - ∫ E .ds

In this case the electric field is in the direction and the points (ds) are also in the direction and therefore the angle is zero and the scalar product is reduced to the algebraic product.

         V_{b} - V_{a} = - ∫ E ds

We substitute

         V_{b} - V_{a} = - ∫ (α + β/ y²) dy

We integrate

          V_{b} - V_{a} = - α y + β / y

We evaluate between the lower limit A  2 cm = 0.02 m and the upper limit B 3 cm = 0.03 m

           V_{b} - V_{a} = - α (0.03 - 0.02) + β (1 / 0.03 - 1 / 0.02)

            V_{b} - V_{a} = - 600 0.01 + 5 (-16.67) = -6 - 83.33

            V_{b} - V_{a} = - 89.3 V

As they ask us the reverse case

             V_{b} - V_{a} = - V_{b} - V_{a}

             V_{a} - V_{b} = 89.3

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When a car drives along a "washboard" road, the regular bumps cause the wheels to oscillate on the springs. (What actually oscil
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Answer:

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part a)

First in order to estimate the spring constant k, we need to know the expression or formula to use in this case:

k = ΔF / Δx

Where:

ΔF: force that the men puts in the car, in this case, the weight.

Δx: the sinking of the car, which is 2 cm or 0.02 m.

With this data, and knowing that there are four mens, replace the data in the above formula:

W = 80 * 10 = 800 N

This is the weight for 1 man, so the 4 men together would be:

W = 800 * 4 = 3200 N

So, replacing this data in the formula:

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To get the frequency, we need to know the angular speed of the car with the following expression:

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Replacing the data:

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And the frequency:

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c) finally for the speed, we have the time and the distance, so:

V = x * t

The only way to hit bumps at this frequency, is covering the gaps of bumping, about 6 times per second so:

x: distance of 80 cm or 0.8 m

V = 0.8 * 6 =

V = 4.8 m/s

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