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3 years ago

Which explains how time of winds creates high waves?

Physics

2 answers:

NeX [460]3 years ago

4 0

C.
Winds blowing for a short time create powerful, high, and fast waves.

lisabon 2012 [21]3 years ago

3 0

C
The smaller waves created by the constant winds gradually add up to form larger ones.

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The rubber band contains .......potential energy as it is stretched.

kodGreya [7K]

Answer:

elastic potential energy

You input potential (stored) energy into the rubber band system when you stretched the rubber band back. Because it is an elastic system, this kind of potential energy is specifically called elastic potential energy.

Explanation:

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1 year ago

How many molecules of Oxygen gas are there on the reactant side of this equation?

Sergeeva-Olga [200]

Answer:

Explanation:

It has 8 O atoms and 4 O2(g) molecules

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2 years ago

You rub a clear plastic pen with wool, and observe that a strip of invisible tape is attacted to the pen. Assuming that the pen

melomori [17]

Explanation:

It is known that opposite charges tend to attract each other whereas like charges tend to repel each other.

For example, when we rub a plastic pen with wool and it gets attracted towards an invisible tape then it means that there is an opposite charge present on both the objects.

A neutral object also tends to develop a charge when it comes in contact with a charged object.

Thus, we can conclude that the following statements are true.

The tape might be positively charged.

The tape might be uncharged.

3 0

2 years ago

Sum of number of protons and neutrons in the nucleus

Alina [70]

That's the "Atomic Mass" of the atom.

4 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

3 years ago