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lys-0071 [83]
4 years ago
11

A wooden ring whose mean diameter is 15.0 cm is wound with a closely spaced toroidal winding of 555 turns. Compute the magnitude

of the magnetic field at the center of the cross section of the windings when the current in the windings is 0.670 A.
Physics
1 answer:
ruslelena [56]4 years ago
7 0

Answer:

9.916\times 10^{-4}T

Explanation:

Diameter of toroid is 15 cm =0.15 m

So length of the toroid is l=\pi d=3.14\times 0.15=0.471\ m

Number of turns of the toriod is given as 555

Current through the toroid is 0.670 A

Magnetic field B=\mu _0ni=\mu _0\frac{N}{L}i=4\pi \times 10^{-7}\times \frac{555}{0.471}\times 0.670=9.916\times 10^{-4}T

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Answer:

The theory is supported by all the available observations and data.

Explanation: The scientific community will accept a theory when a sufficient body of evidence supports it. This includes experiments that refute other potential theories. Experiments should also be carried out that attempt to disprove the theory but cannot.

It should not matter who proposed the theory or who supports it, and instead should entirely be based on the quality and abundance of data supporting it.

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3 years ago
Type your answers to these 4 questions in the submission area.
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What type of image is formed by a lens if m = -1.6?
anastassius [24]

The image formed from the lens is real, inverted and enlarged

Explanation:

The magnification of a lens is given by:

m=\frac{y'}{y}=-\frac{q}{p} (1)

where:

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y is the size of the object

q is the distance of the image from the lens

p is the distance of the object from the lens

For this lens, the magnification is

m = -1.6

We notice the following:

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Looking at eq.(1), we can therefore make the following conclusions:

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5 0
3 years ago
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These types of electromagnetic waves are right next to red light on the electromagnetic spectrum:
vodomira [7]
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4 years ago
A horizontal spring with a 10000 N/m spring constant is compressed 0.08 m, and a 12-kg block is placed against it. When the bloc
agasfer [191]

Answer:

4.04m

Explanation:

First you calculate the velocity of the block when it leaves the spring. You calculate this velocity by taking into account that the potential energy of the spring equals the kinetic energy of the block, that is:

U=K\\\\\frac{1}{2}kx^2=\frac{1}{2}mv^2\\\\v=\sqrt{\frac{kx^2}{m}}=\sqrt{\frac{(10000N/m)(0.08m)^2}{12kg}}=2.3\frac{m}{s}

To find the distance in which the block stops you use the following expression (net work done by the friction force is equal to the difference in the kinetic energy of the block):

W_{T}=\Delta K\\\\F_f d=\frac{1}{2}m[v^2-v_o^2]\\\\d=\frac{mv^2}{2F_f}

where Ff is the friction force. By replacing the values of the parameters you obtain:

d=\frac{(12kg)(2.3m/s)^2}{2(8N)}=3.96m

hence, the distance to the original position is 3.96m+0.08m=4.04m

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3 years ago
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