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Natalka [10]
3 years ago
9

If the magnetic field strength is found to be zero between the two wires at a distance of 3.0 cm from the first wire, what is th

e magnitude of the current in the second wire?

Physics
1 answer:
Taya2010 [7]3 years ago
4 0

Answer: The magnitude of the current in the second wire 2.67A

Explanation:

Here is the complete question:

Two straight parallel wires are separated by 7.0 cm. There is a 2.0-A current flowing in the first wire. If the magnetic field strength is found to be zero between the two wires at a distance of 3.0 cm from the first wire, what is the magnitude of the current in the second wire?

Explanation: Please see the attachments below

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A string with a length of 4.00 m is held under a constant tension. The string has a linear mass density of \mu=0.000600~\text{kg
yulyashka [42]

Answer:

T=245.76N

Explanation:

We know that the frequency of the nth harmonic is given by f_n=nf, where f is the fundamental harmonic. Since we have the values of two consecutive frequencies, we can do:

f_{n+1}-f_n=(n+1)f-nf=nf+f-nf=f

Which for our values means (we do not need the value of <em>n</em>, that is, which harmonics are the frequencies given):

f=f_{n+1}-f_n=480Hz-400Hz=80Hz

Now we turn to the formula for the vibration frequency of a string (for the fundamental harmonic):

f=\frac{1}{2L} \sqrt{\frac{T}{\mu}}

So the tension is:

T=\mu(2Lf)^2

Which for our values is:

T=(0.0006kg/m)(2(4m)(80Hz))^2=245.76N

6 0
3 years ago
An ice skater has a moment of inertia of 5.0 kgm2 when her arms are outstretched. at this time she is spinning at 3.0 revolution
givi [52]
With arms outstretched,
Moment of inertia is I = 5.0 kg-m².
Rotational speed is ω = (3 rev/s)*(2π rad/rev) = 6π rad/s
The torque required is
T = Iω = (5.0 kg-m²)*(6π rad/s) = 30π 

Assume that the same torque drives the rotational motion at a moment of inertia of 2.0 kg-m².
If u = new rotational speed (rad/s), then
T = 2u = 30π
u = 15π rad/s
   = (15π rad/s)*(1 rev/2π rad)
   = 7.5 rev/s

Answer: 7.5  revolutions per second.

7 0
3 years ago
The gold foil experiment led to the conclusion that each atom in the foil was composed mostly of empty space because most alpha
katovenus [111]

Answer:

(1) passed through the foil

Explanation:

Ernest Rutherford conducted an experiment using an alpha particle emitter projected towards a gold foil and the gold foil was surrounded by a fluorescent screen which glows upon being struck by an alpha particle.

  • When the experiment was conducted he found that most of the alpha particles went away without any deflection (due to the empty space) glowing the fluorescent screen right at the point of from where they were emitted.
  • While a few were deflected at reflex angle because they were directed towards the center of the nucleus having the net effective charge as positive.
  • And some were acutely deflected due to the field effect of the positive charge of the proton inside the nucleus. All these  conclusions were made based upon the spot of glow on the fluorescent screen.

5 0
3 years ago
Read 2 more answers
How could you increase the kinetic energy of a wagon without increasing it's mass
Zarrin [17]

Answer:

send the wagon down a higher hill

3 0
3 years ago
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What is the restoring force of a pendulum
salantis [7]

Answer:

the Restoring force causes the vibrating object to go slower going further from the equilibrium position and to go faster as it approaches the equilibrium position. the restoring force is what is causing the vibration The tension force comes from the string tugging on the bob of the pendulum.

Explanation:

3 0
2 years ago
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