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kiruha [24]
3 years ago
8

If a bar of copper is brought near a magnet, the copper bar will be

Physics
2 answers:
STALIN [3.7K]3 years ago
8 0

Answer: Option B

Explanation: Copper is a non-ferrous material (like aluminuim for example), and the only materials affected by a magnet (or a magnetic field) are the ferrous ones, so the correct answer will be the option B, the copper will be unaffected by the magnet.

Alecsey [184]3 years ago
5 0
It will be unaffected by the magnet because it has no magnetic field. If you were to maybe have electricity going through it is the only way it would have anything to do with the magnet. 
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Two facing surfaces of two large parallel conducting plates separated by 8.5 cm have uniform surface charge densities such that
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Answer:

positive plate

E = 5.764 KV / m

W = 490eV or 7.85 * 10^-17 J

E_p = 4.74 *10^(-12) eV

E_k = 490 eV

Explanation:

part a

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Distance between two plates = 8.5 cm

Answer: The positive plate is at higher potential because of convention.

part b

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E = V / d

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Answer: Electric Field between the plates E = 5.764 KV / m

part c

Work done by electric field

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W = 7.85 * 10^-17 J

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E_p = m_e * g * d / (1.602*10^-19)

E_p =  9.109*10^-31* 9.81 * 0.085 / (1.602*10^-19)

E_p = 4.74 *10^(-12) eV

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part e

Kinetic Energy of an electron gained:

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E_k = 490eV - 4.74*10^(-12)eV

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7 0
3 years ago
Sound travels through air at 343 m/s,
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The sound wave will have traveled 2565 m  farther in water than in air.

Answer:

Explanation:

It is known that distance covered by any object is directly proportional to the velocity of the object and the time taken to cover that distance.

Distance = Velocity × Time.

So if time is kept constant, then the distance covered by a wave can vary depending on the velocity of the wave.

As we can see in the present case, the velocity of sound wave in air is 343 m/s. So in 2.25 s, the sound wave will be able to cover the distance as shown below.

Distance = 343 × 2.25 =771.75 m

And for the sound wave travelling in fresh water, the velocity is given as 1483 m/s. So in a time interval of 2.25 s, the distance can be determined as the product of velocity and time.

Distance = 1483×2.25=3337 m.

Since, the velocity of sound wave travelling in fresh water is greater than the sound wave travelling in air, the distance traveled by sound wave in fresh water will be greater.

Difference in distance covered in water and air = 3337-772 m = 2565 m

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