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mestny [16]
3 years ago
5

A magnet falls through a loop of wire with the south pole entering first. After it has fallen all the way through the wire loop

so that the north pole end is moving away from the loop, what is the direction of the induced current in the loop (as viewed from above, looking down on the magnet)
Physics
1 answer:
Alisiya [41]3 years ago
5 0

Answer:

The induced current direction as viewed is clockwise

Explanation:

Lenz's Law states that the induced e. m. f. causes current to be driven in the loop of wire in such a way as to generate magnetic field that are oppose the magnetic flux change which is the source of the induced current

Therefore, as the magnet approaches the coil with the south pole,  the coil produces current equivalent to the upward movement of the south pole of a permanent magnet through it which according to Flemings Right Hand Rule is clockwise

Therefore;

The direction of the induced current in the loop (as viewed from above, looking down the magnet) is clockwise

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Which refers to any disturbance that carries energy from one place to another through matter and space?
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I think frequency it sounds like the correct answer but I am not completely sure if I am correct
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When a piano tuner strikes both the A on the piano and a 440 Hz tuning fork, he hears a beat every 2 seconds. The frequency of t
AysviL [449]

Answer:

The  frequency is  f = 439.5 \  Hz        

Explanation:

From the question we are told that

   The frequency of the  tuning fork is  f_t  =  440 \ Hz

   The beat period is  T  =  2 \  s

Generally the beat frequency is mathematically represented as

       f_b  =  \frac{1}{T}

       f_b  =  \frac{1}{2}

      f_b  = 0.5 \  Hz

The  beat frequency is also represented mathematically as

     f_b  =  f_t  -  f

Where  f is the frequency of the piano

 So

       f =  440 -  0.5  

       f = 439.5 \  Hz        

3 0
3 years ago
Question 2 of 10
Mekhanik [1.2K]

the awnser to ur question is D

6 0
3 years ago
Consider a container of oxygen gas at a temperature of 23°C that is 1.00 m tall. Compare the gravitational potential energy of a
Sergio039 [100]

Answer:

Yes, it is reasonable to neglect it.

Explanation:

Hello,

In this case, a single molecule of oxygen weights 32 g (diatomic oxygen) thus, the mass of kilograms is (consider Avogadro's number):

m=1molec*\frac{1mol}{6.022x10^{23}molec} *\frac{32g}{1mol}*\frac{1kg}{1000g}=5.31x10^{-26}kg

After that, we compute the potential energy 1.00 m above the reference point:

U=mhg=5.31x10^{-26}kg*1.00m*9.8\frac{m}{s^2}=5.2x10^{-25}J

Then, we compute the average kinetic energy at the specified temperature:

K=\frac{3}{2}\frac{R}{Na}T

Whereas N_A stands for the Avogadro's number for which we have:

K=\frac{3}{2} \frac{8.314\frac{J}{mol*K}}{6.022x10^{23}/mol}*(23+273)K\\ \\K=6.13x10^{-21}J

In such a way, since the average kinetic energy energy is about 12000 times higher than the potential energy, it turns out reasonable to neglect the potential energy.

Regards.

8 0
3 years ago
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