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Delvig [45]
3 years ago
9

A bar magnet is held above the center of a conducting ring in the horizontal plane. The magnet is dropped so it falls lengthwise

toward the center of the ring. Will the falling magnet be attracted toward the ring or be repelled by the ring due to the magnetic interaction of the magnet and the ring?
Physics
1 answer:
Alenkinab [10]3 years ago
7 0

Explanation:

Since, it is given that the magnet drops and falls lengthwise towards the canter of the ring. As a result, change in magnetic flux will occur which tends to induce an electric current in the ring.

Therefore, a magnetic field is also produced by the ring itself which will actually oppose or repel the magnet.  

Thus, we can conclude that the falling magnet be repelled by the ring due to the magnetic interaction of the magnet and the ring.

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What happens to the average kinetic energy of water molecules as water freezes? It decreases as the water absorbs energy from it
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Answer:the average kinetic energy is directly proportional to the temperature . When water freezes, the temperature decreases and therefore, the average kinetic energy will also decreases as well.

Explanation:

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The collision between a hammer and a nail can be considered to be approximately elastic. estimate the kinetic energy acquired by
Setler [38]

Here we can use momentum conservation as in this type of collision there is no external force on it

m_1v_{1i} + m_2v_{2i} = m_1 v_{1f} + m_2v_{2f}

now here we can say

m_1 = 10 g

v_{1i} = 0

m_2 = 550 g

v_{2i} = 3.5 m/s

now here we can say

10*0 + 550 * 3.5 = 10 v_{1f} + 550 v_{2f}

192.5 = v_{1f} + 55 v_{2f}

now by coefficient of restitution

for elastic collision we know that e = 1

v_{2f} - v_{1f} = e(v_{1i} - v_{2i})

v_{2f} - v_{1f} = 0 - 3.5

now by solving the two equation

56v_{2f} = 189

v_{2f} = 3.375 m/s

also we know that

v_{1f} = v_{2f} + 3.5 = 3.375 + 3.5 = 6.875 m/s

so final speed of the nail is 6.875 m/s


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A piano wire with mass 2.95 g and length 79.0 cm is stretched with a tension of 29.0 N . A wave with frequency 105 Hz and amplit
Romashka-Z-Leto [24]

The concept needed to solve this problem is average power dissipated by a wave on a string. This expression ca be defined as

P = \frac{1}{2} \mu \omega^2 A^2 v

Here,

\mu = Linear mass density of the string

\omega =  Angular frequency of the wave on the string

A = Amplitude of the wave

v = Speed of the wave

At the same time each of this terms have its own definition, i.e,

v = \sqrt{\frac{T}{\mu}} \rightarrow Here T is the Period

For the linear mass density we have that

\mu = \frac{m}{l}

And the angular frequency can be written as

\omega = 2\pi f

Replacing this terms and the first equation we have that

P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2(\sqrt{\frac{T}{\mu}})

P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2 (\sqrt{\frac{T}{m/l}})

P = 2\pi^2 f^2A^2(\sqrt{T(m/l)})

PART A ) Replacing our values here we have that

P = 2\pi^2 (105)^2(1.8*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})

P = 0.2320W

PART B) The new amplitude A' that is half ot the wavelength of the wave is

A' = \frac{1.8*10^{-3}}{2}

A' = 0.9*10^{-3}

Replacing at the equation of power we have that

P = 2\pi^2 (105)^2(0.9*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})

P = 0.058W

8 0
2 years ago
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