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Lisa [10]
4 years ago
11

15.Restore the battery setting to 10 V. Now change the number of loops from 4 to 3. Explain what happens to the magnitude and di

rection of the magnetic field. Now change to 2 loops, then to 1 loop. What do you observe the relationship to be between the magnitude of the magnetic field and the number of loops for the same current
Physics
1 answer:
lozanna [386]4 years ago
6 0

Answer:

we see it is a linear relationship.

Explanation:

The magnetic flux is u solenoid is

      B = μ₀ N/L   I

where N is the number of loops, L the length and I the current

By applying this expression to our case we have that the current is the same in all cases and we can assume the constant length. Consequently we see that the magnitude of the magnetic field decreases with the number of loops

      B = (μ₀ I / L)  N

the amount between paracentesis constant, in the case of 4 loop the field is worth

      B = cte 4

N       B

4       4 cte

3       3 cte

2       2 cte

1        1 cte

as we see it is a linear relationship.

In addition, this effect for such a small number of turns the direction of the field that is parallel to the normal of the lines will oscillate,

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igomit [66]

Answer: Linear speed is 1,670 Kph.

Explanation:

If we assume that the earth is a perfect sphere, and that is spinning itself once every roughly 24 hr, we can get the angular velocity of the Earth, in magnitude, as follows:

ω = 2π / 24 Hr

Now, by definition, an angle is the relationship between the arc s, and the radius r, so we can replace these values in the angular velocity expression, as follows:

ω = (Δs / r) . 1/Δt ⇒ ω = (Δs/Δt). 1/r

But, by definition, Δs/At, is just the linear velocity, v, so we can conclude the following;

ω = v/r ⇒ v = ω. r

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v = 2π /24 hr . 6378 Km = 1,670 Km/hr.

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3 years ago
Who clarified the photoelectric effect?
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4 years ago
A wave travels at 295 m/s and has a wavelength of 2.50 m. What is the frequency of the wave?
posledela

Answer:

118\; \rm Hz.

Explanation:

The frequency f of a wave is equal to the number of wave cycles that go through a point on its path in unit time (where "unit time" is typically equal to one second.)

The wave in this question travels at a speed of v= 295\; \rm m\cdot s^{-1}. In other words, the wave would have traveled 295\; \rm m in each second. Consider a point on the path of this wave. If a peak was initially at that point, in one second that peak would be

How many wave cycles can fit into that 295\; \rm m? The wavelength of this wave\lambda = 2.50\; \rm m gives the length of one wave cycle. Therefore:

\displaystyle \frac{295\;\rm m}{2.50\; \rm m} = 118.

That is: there are 118 wave cycles in 295\; \rm m of this wave.

On the other hand, Because that 295\; \rm m of this wave goes through that point in each second, that 118 wave cycles will go through that point in the same amount of time. Hence, the frequency of this wave would be

Because one wave cycle per second is equivalent to one Hertz, the frequency of this wave can be written as:

f = 118\; \rm s^{-1} = 118\; \rm Hz.

The calculations above can be expressed with the formula:

\displaystyle f = \frac{v}{\lambda},

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