Answer:
Explanation:
The magnetic field due to straight wire is into the square coil.
As the current in straight wire decreases the magnetic flux in the coil decreases
. The induced magnetic field is into the coil.The induced current is along +y direction
Answer:
O²⁻
Explanation:
Number of protons = 8
Number of neutrons = 9
Number of electrons = 10
What type of atom or ion is it = ?
Solution:
Protons are the positively charged particle in an atom
Neutrons do not carry any charges
Electrons are negatively charged particles
For this atom, the number of protons helps to identify what specie it is; so this is an oxygen atom.
Now,
Charge = Number of protons - Number of electrons
Charge = 8 - 10 = -2
The charge on the atom is -2 and so it is an oxygen ion with -2 charge
The ion is O²⁻
Answer:
10 N
Explanation:
F = ma = m(Δv/t) = 5.0(10.0 - 0)/5.0 = 10 N
Answer:
Recall that the electric field outside a uniformly charged solid sphere is exactly the same as if the charge were all at a point in the centre of the sphere:

lnside the sphere, the electric field also acts like a point charge, but only for the proportion of the charge further inside than the point r:

To find the potential, we integrate the electric field on a path from infinity (where of course, we take the direct path so that we can write the it as a 1 D integral):

=![\frac{q}{4\pi e_{0} } [\frac{1}{R} -\frac{r^{2}-R^{2} }{2R^{3} } ]](https://tex.z-dn.net/?f=%5Cfrac%7Bq%7D%7B4%5Cpi%20e_%7B0%7D%20%7D%20%5B%5Cfrac%7B1%7D%7BR%7D%20-%5Cfrac%7Br%5E%7B2%7D-R%5E%7B2%7D%20%20%7D%7B2R%5E%7B3%7D%20%7D%20%5D)
∴NOTE: Graph is attached
m=23.8kg a=8.97m/s^2 Fnet=? Fnet=ma=(23.8kg)(8.97m/s^2)=213.486N