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Goryan [66]
3 years ago
6

For the parallel plates mentioned above, the DC power supply is set to 31.5 Volts and the plate on the right is at x = 14 cm. Wh

at is the magnitude of the electric field at a point on the x-axis where x = 7.0 cm? Answer with a number in the format ### in Newtons per Coulombs.

Physics
1 answer:
Aleonysh [2.5K]3 years ago
8 0

Note: The complete question is attached as a file to this solution. The parallel plate mentioned can be seen in this picture attached.

Answer:

E = 225 N/C

Explanation:

Note: At any point on the parallel plates of a capacitor, the electric field is uniform and equal.

Therefore, Electric field at x = 14 cm equals the electric field at x = 7 cm

V(x) = 31.5 Volts

x = 14 cm = 0.14 m

The magnitude of the electric field at any point between the parallel plate of the capacitor is given by the equation:

E = V(x)/d

E(x = 0.14) = 31.5/0.14

E(x=0.14) = 225 N/C

E(x=0.14) = E(x=0.07) = 225 N/C

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Answer:

The center of mass of the Earth-Moon system is 4.673 kilometers away from center of Earth.

Explanation:

Let suppose that planet and satellite can be treated as particles. The masses of Earth and Moon (m_{E}, m_{M}) are 5.972\times 10^{24}\,kg and 7.349\times 10^{22}\,kg, respectively. The distance between centers is 384,403 kilometers. The location of the center of mass can be found by using weighted averages:

\bar x = \frac{x_{E}\cdot m_{E}+x_{M}\cdot m_{M}}{m_{E}+m_{M}}

If x_{E} = 0\,km and x_{M} = 384,403\,km, then:

\bar x = \frac{(0\,km)\cdot (5.972\times 10^{24}\,kg)+(384,403\,km)\cdot (7.349\times 10^{22}\,kg)}{5.972\times 10^{24}\,kg+7.349\times 10^{22}\,kg}

\bar x = 4.673\,km

The center of mass of the Earth-Moon system is 4.673 kilometers away from center of Earth.

8 0
3 years ago
We can expect the force of friction to be greater for an object moving on a surface if the surfaces in contact are
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Answer:

c

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8 0
2 years ago
The river narrows at a rapids from a width of 12 m to a width of only 5.8 m. The depth of the river before the rapids is 2.7 m;
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Answer:

7.89 m/s

Explanation:

Given that

Width of the river, b1 = 12 m

Width of the river, b2 = 5.8 m

Depth of the river, d1 = 2.7 m

Depth of the river, d2 = 0.85 m

Speed of the river, v1 = 1.2 m/s

Speed of the river, v2 = ?

Area of the river before the rapid, a1 = 12 * 2.7 = 32.4 m²

Area of the river after the rapid, a2 = 5.8 * 0.85 = 4.93 m²

To solve this question, we use a relation between the speed of the river and the volume of the river. We say,

Area1 * velocity1 = Area2 * velocity2, and when we substitute the values for each other we have

32.4 * 1.2 = 4.93 * v2

38.88 = 4.93v2

v2 = 38.88 / 4.93

v2 = 7.89 m/s

Therefore, the speed of the river after the rapid is 7.89 m/s

6 0
3 years ago
What is my name if it starts with a then m then a then r then y?
Ierofanga [76]

Answer:

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Explanation:

?

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Lucas left a metal bowl and a wooden bowl outside overnight. The next morning, he picked up the bowls to bring them inside. He n
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Answer: C. Metal transfers heat away from the skin by conduction, creating the sensation of coolness.

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