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3 years ago

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An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm2, separated by a distance of 1.70 mm. A 21

.0-V potential difference is applied to these plates. (a) Calculate the electric field between the plates.

1 answer:

Inga [223]3 years ago

8 0

Answer:

12353 V m⁻¹ = 12.4 kV m⁻¹

Explanation:

Electric field between the plates of the parallel plate capacitor depends on the potential difference across the plates and their distance of separation.Potential difference across the plates V over the distance between the plates gives the electric field between the plates. Potential difference is the amount of work done per unit charge and is given here as 21 V. Electric field is the voltage over distance.

E = V ÷ d = 21 ÷ 0.0017 = 12353 V m⁻¹

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A potter's wheel with rotational inertia 7.5 kg m2 is spinning freely at 16.0 rpm. The potter drops a 2.65 kg lump of clay on th

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4 years ago

A car is stationary. It accelerates at 0.8 ms^2 for 10 s and then at 0.4 ms^2 for a further 10 s. Use the equations of motion to

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Explanation:

The final displacement of the car ( $s$ ), in meters, is the sum of the change in displacement associated with each part of the journey, which is derived from the following kinematic formulas:

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$s_{2} = v_{o,2} + \frac{1}{2}\cdot a_{2}\cdot t_{2}^{2}$ (4)

Where:

$s_{1}$ - Traveled distance of the first part, in meters.

$s_{2}$ - Traveled distance of the second part, in meters.

$a_{1}$ - Acceleration in the first part, in meters per square second.

$a_{2}$ - Acceleration in the second part, in meters per square second.

$v_{o,2}$ - Initial speed of the car in the second part, in meters per second.

$t_{1}$ - Time taken in the first part, in seconds.

$t_{2}$ - Time taken in the second part, in seconds.

If we know that $a_{1} = 0.8\,\frac{m}{s^{2}}$ , $t_{1} = 10\,s$ , $a_{2} = 0.4\,\frac{m}{s^{2}}$ and $t_{2} = 10\,s$ , then the distance traveled by the car is:

By (2):

$s_{1} = \frac{1}{2}\cdot \left(0.8\,\frac{m}{s^{2}} \right)\cdot(10\,s)^{2}$

$s_{1} = 40\,m$

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$v_{o,2} = \left(0.8\,\frac{m}{s^{2}} \right)\cdot (10\,s)$

$v_{o,2} = 8\,\frac{m}{s}$

By (4):

$s_{2} = \left(8\,\frac{m}{s} \right)\cdot (10\,s) + \frac{1}{2}\cdot \left(0.4\,\frac{m}{s^{2}} \right)\cdot (10\,s)^{2}$

$s_{2} = 100\,m$

By (1):

$s = 140\,m$

The final displacement of the car is 140 meters.

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