Answer:
22 N applied force
Explanation: Since they are both pushing the wagon in the same direction the force adds up.
(a). If the temperature of a substance is increased the density of it will also increased. The temperature is directly proportional to the density because the molecular activity of the substance will be affected by the temperature
(b).The mass of the substance doesn't at all. Because the amount of matter inside the substance cannot be affected by the temperature except that its water
Explanation:
In the given question, the two metal spheres were hanged with the nylon thread.
When these two spheres were brought close together, they attracted each other. The attraction between these spheres is the result of the opposite charges between them.
The possible ways by which these two metal spheres can be charged are by induction that is touching the metal or by rubbing them.
During induction, the same charges are transferred to each sphere. In this case, either both the spheres will be negatively charged or positively charged.
It is not possible that after the sphere touch each other they will cling together because the same charge repels each other and during touching, if one sphere is neutral, then the charged one will transfer the same charge. And as we know that same charge repel each other therefore they will repel each other.
Answer:
![B_T=2.0*10^-5[-\hat{i}+\hat{j}]T](https://tex.z-dn.net/?f=B_T%3D2.0%2A10%5E-5%5B-%5Chat%7Bi%7D%2B%5Chat%7Bj%7D%5DT)
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:

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}]](https://tex.z-dn.net/?f=B_T%3DB_1%2BB_2%2BB_3%5C%5C%5C%5CB_%7BT%7D%3D%5Cfrac%7B%5Cmu_o%20I_1%7D%7B2%5Cpi%20r_1%7D%5Chat%7Bj%7D-%5Cfrac%7B%5Cmu_o%20I_2%7D%7B2%5Cpi%20r_2%7D%5Chat%7Bi%7D%2B%5Cfrac%7B%5Cmu_o%20I_3%7D%7B2%5Cpi%20r_3%7D%5B-cos45%5Chat%7Bi%7D%2Bsin45%5Chat%7Bj%7D%5D)
I1 = I2 = I3 = 6.0A
r1 = 0.09m
r2 = 0.09m

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](https://tex.z-dn.net/?f=B_T%3D%5Cfrac%7B%5Cmu_o%20I%7D%7B2%5Cpi%7D%5B%28-%5Cfrac%7B1%7D%7Br_2%7D-%5Cfrac%7Bcos45%7D%7Br_3%7D%29%5Chat%7Bi%7D%2B%28%5Cfrac%7B1%7D%7Br_1%7D%2B%5Cfrac%7Bsin45%7D%7Br_3%7D%29%5Chat%7Bj%7D%7D%5D%5C%5C%5C%5CB_T%3D%5Cfrac%7B%284%5Cpi%2A10%5E%7B-7%7DT%2FA%29%286.0A%29%7D%7B2%5Cpi%7D%5B%28-%5Cfrac%7B1%7D%7B0.09m%7D-%5Cfrac%7Bcos45%7D%7B0.127m%7D%29%5Chat%7Bi%7D%2B%28%5Cfrac%7B1%7D%7B0.09m%7D%2B%5Cfrac%7Bsin45%7D%7B0.127m%7D%29%5D%5C%5C%5C%5CB_T%3D%5Cfrac%7B%284%5Cpi%2A10%5E%7B-7%7DT%2FA%29%286.0A%29%7D%7B2%5Cpi%7D%5B-16.67%5Chat%7Bi%7D%2B16.67%5Chat%7Bj%7D%5D%5C%5C%5C%5CB_T%3D2.0%2A10%5E-5%5B-%5Chat%7Bi%7D%2B%5Chat%7Bj%7D%5DT)