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RSB [31]
2 years ago
6

Please answer this science question NO LINKS !!!!

Physics
1 answer:
Sergeu [11.5K]2 years ago
4 0
FNet = Fa + Fg. Is the formula. I think the 20 should be right
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Ksenya-84 [330]

I am a girl and i have a virgina and im 13 tho so dont touch me lolz mark brainly

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2 years ago
How many ounces is 200 lbs converted
ddd [48]

Answer:

3,200 ounces

Explanation:

1 pound (lb) is equal to 16 ounces (oz):

i.e 1 lb = 16 oz

Given:

200 pounds

To find:

2,000 pounds as ounces.

Steps:

200 (mass) * 16 = 3,200 ounces.

Thank you!

- EE

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2 years ago
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A total electric charge of 4.00 nC is distributed uniformly over the surface of a metal sphere with a radius of 26.0 cm. The pot
nevsk [136]

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7 0
1 year ago
A guitar string is plucked. What best describes the formation and transmission of the sound wave?
makvit [3.9K]

The string vibrates the surrounding air, which produces a longitudinal wave.

Brainly if correct? Please and thank you :D

6 0
3 years ago
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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
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