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Bumek [7]
3 years ago
12

A person standing in the moon's "blank" would see a total solar eclipse ​

Physics
2 answers:
MakcuM [25]3 years ago
6 0

Answer:

umbra

Explanation:

Tatiana [17]3 years ago
5 0

Answer:

<h2>e3r1</h2>

Explanation:

ef;ggfhhfjtjftftccth

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What happens if :<br> . The test charge is not tiny.
docker41 [41]

The magnitude of the test charge must be small enough so that it does not disturb the issuance of the charges whose electric field we wish to measure otherwise the metric field will be different from the actual field.

<h3>How does test charge affect electric field?</h3>

As the quantity of authority on the test charge (q) is increased, the force exerted on it is improved by the same factor. Thus, the ratio of force per charge (F / q) stays the same.

Adjusting the amount of charge on the test charge will not change the electric field force.

<h3>What is a test charge used for?</h3>

The charge that is used to measure the electric field strength is directed to as a test charge since it is used to test the field strength. The test charge has a portion of charge denoted by the symbol q.

To learn more about test charge, refer

brainly.com/question/16737526

#SPJ9

3 0
2 years ago
PLEASE HELP!!!!
Alex_Xolod [135]
I would say A Because it weighs more than the water
5 0
3 years ago
A cube-shaped piece of copper has sides of 4cm each and it's density is
iris [78.8K]

Answer:

64 cm^3

Explanation:

<u>Density </u>

The density of a substance is the mass per unit volume. The density varies with temperature and pressure.

The formula to calculate the density of a substance of mass (m) and volume (V) is:

\displaystyle \rho=\frac{m}{V}

We have a cube-shaped piece of copper of 4 cm of side length. The volume of the piece is:

V=(4\ cm)^3=64\ cm^3

Surprisingly, no other magnitude is required, thus the answer is:

\mathbf{64 cm^3}

5 0
3 years ago
How many neutrons are present in an atom of Iron-54
Flura [38]

there are 28 neutrons present

3 0
3 years ago
Four long wires are each carrying 6.0 A. The wires are located
Firdavs [7]

Answer:

B_T=2.0*10^-5[-\hat{i}+\hat{j}]T

Explanation:

To find the magnitude of the magnetic field, you use the following formula for the calculation of the magnetic field generated by a current in a wire:

B=\frac{\mu_oI}{2\pi r}

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

I: current = 6.0 A

r: distance to the wire in which magnetic field is measured

In this case, you have four wires at corners of a square of length 9.0cm = 0.09m

You calculate the magnetic field in one corner. Then, you have to sum the contribution of all magnetic field generated by the other three wires, in the other corners. Furthermore, you have to take into account the direction of such magnetic fields. The direction of the magnetic field is given by the right-hand side rule.

If you assume that the magnetic field is measured in the up-right corner of the square, the wire to the left generates a magnetic field (in the corner in which you measure B) with direction upward (+ j), the wire down (down-right) generates a magnetic field with direction to the left (- i)  and the third wire generates a magnetic field with a direction that is 45° over the horizontal in the left direction (you can notice that in the image attached below). The total magnetic field will be:

B_T=B_1+B_2+B_3\\\\B_{T}=\frac{\mu_o I_1}{2\pi r_1}\hat{j}-\frac{\mu_o I_2}{2\pi r_2}\hat{i}+\frac{\mu_o I_3}{2\pi r_3}[-cos45\hat{i}+sin45\hat{j}]

I1 = I2 = I3 = 6.0A

r1 = 0.09m

r2 = 0.09m

r_3=\sqrt{(0.09)^2+(0.09)^2}m=0.127m

Then you have:

B_T=\frac{\mu_o I}{2\pi}[(-\frac{1}{r_2}-\frac{cos45}{r_3})\hat{i}+(\frac{1}{r_1}+\frac{sin45}{r_3})\hat{j}}]\\\\B_T=\frac{(4\pi*10^{-7}T/A)(6.0A)}{2\pi}[(-\frac{1}{0.09m}-\frac{cos45}{0.127m})\hat{i}+(\frac{1}{0.09m}+\frac{sin45}{0.127m})]\\\\B_T=\frac{(4\pi*10^{-7}T/A)(6.0A)}{2\pi}[-16.67\hat{i}+16.67\hat{j}]\\\\B_T=2.0*10^-5[-\hat{i}+\hat{j}]T

5 0
4 years ago
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