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katrin [286]
2 years ago
5

if there is a duck behind two ducks and a duck between two ducks and a duck in front of two ducks how many ducks are there? look

carefully.....
Physics
1 answer:
TiliK225 [7]2 years ago
8 0
Five ducks I guess so
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A force of 5N compresses a spring by 4cm. Find the force constant of the spring.​
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Answer:

k = 1,250 N/m

Explanation:

Use the formula F=kx, with F=5N and x=0.04m

Then the spring constant (k) is 5/0.04

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(iv) Rank the capacitors in part (i) in the order of the energy they store.
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5 0
1 year 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
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