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sp2606 [1]
3 years ago
7

A neutral metal comb is held near an object with a negative charge. What happens to the comb?

Physics
1 answer:
mixas84 [53]3 years ago
5 0

Answer:

Neutral comb is charged through induction.

Explanation:

The local charging of the comb is caused because of induction. The electron is pushed in the comb to the away side of the charged particle because of the negative charge which makes the close part positively charge.

Then after that when the observer move the comb away from the charge then the electron of the comb is redistributed.

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Which of Jafar's statements are correct regarding distance and displacement?
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Answer:The distance and magnitude of displacement are sometimes equal." Jafar is correct. The distance traveled and the magnitude of displacement are equal if and only if the path is a straight line in one direction.

Explanation:

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___ + 3H2O + light —> C3H6O3 + 3O2. What amount and substance balance this reaction?
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Answer:3H2O + light-c3h603+302

Explanation:

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When light passes from a faster medium into a slower medium, which of the following explains what will occur?
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When light passes from a faster medium into a slower medium, light will be refracted toward a line drawn perpendicular to the point of refraction. <em>(B)</em>

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A copper wire has a length of 1.50 m and a cross sectional area of 0.380 mm2. If the resistivity of copper is 1.70 ✕ 10−8 Ω · m
sweet [91]

Answer: 1.044 E -17 A

Explanation:

L =1.50m

A = 0.380mm = 3.8E- 7

Resistivity p =1.70 E-8

R =pl/A

R = 1.70 E-8 × 1.5/ 3.8 E-7

R = 6.71 E-16 ohms

V = IR

0.700 = I × 6.71 E-16

I =1.044E-17 A

5 0
3 years ago
A cylinder of mass mm is free to slide in a vertical tube. The kinetic friction force between the cylinder and the walls of the
Zinaida [17]

Answer:

y = \frac{-f +/- \sqrt{f^{2} +2kmg}}{k}

Explanation:

Let y₀ be the initial position of the cylinder when the spring is attached and y its position when it is momentarily at rest.From work-kinetic energy principles,  The work done by the spring force + work done by friction + work done by gravity = kinetic energy change of the cylinder

work done by the spring force = ¹/₂k(y₀² - y²)

work done by friction = - f(y - y₀)

work done by gravity = mg(y - y₀)

kinetic energy change of the cylinder = ¹/₂m(v₁² - v₀²)

So ¹/₂k(y₀² - y²) - f(y - y₀) + mg(y - y₀) = ¹/₂m(v₁² - v₀²)

Since the cylinder starts at rest, v₀ = 0. Also, when it is momentarily at rest, v₁ = 0

¹/₂k(y₀² - y²) - f(y - y₀) + mg(y - y₀) = ¹/₂m(0² - 0²)

¹/₂k(y₀² - y²) - f(y - y₀) + mg(y - y₀) = 0

¹/₂ky₀² + fy₀ - mgy₀ -¹/₂ky² - fy + mgy = 0

¹/₂ky₀² + fy₀ - mgy₀ = ¹/₂ky² + fy - mgy

Let y₀ = 0, then the left hand side of the equation equals zero. So,

0 = ¹/₂ky² + fy - mgy

¹/₂ky² + fy - mgy = 0

Using the quadratic formula

y = \frac{-f +/- \sqrt{f^{2} - 4 X\frac{k}{2} X -mg}}{2 X \frac{k}{2} }\\ y = \frac{-f +/- \sqrt{f^{2} +2kmg}}{k}

4 0
3 years ago
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