Answer:

Explanation:
I'll upload my work shortly as an attachment, but here is my process in words:
- In our first situation we have two capacitors in parallel, which means the charge distribution on both of them is the same. With that, we can find a ratio between the values of Capacitors A and B.
- In our second situation , we add a capacitor parallel to A (I called it C). Because A and C are in parallel, we know that they must have the same potential difference; which should come to be 10V since 90V of the total 100V is on B. Also, the equivalent charge distribution across A and C must be equal to that of the charge distribution at B, because A&C are in series with B. So I added the charges on A&C and set that equal to the charge on B.
- Next, I used the ratio from the first situation to substitute Capacitor A out of the equation. This allows us to solve for B's capacitance. (Note: You could have also substituted B for A and solved for A first if you wanted to.)
- Finally, I used B's capacitance to plug back into the ratio from the first situation to find A's capacitance. And they wanted the answer in micro-Farads, so I went ahead and converted each answer to micro.

Magnetic domains are which of the following?
Answer:
6.85 m/s
Explanation:
We can solve the problem by using the law of conservation of momentum.
In fact, since there are no external forces acting, the total momentum before and after must be conserved. So we can write:

where
is the initial mass of the car
is the initial speed of the car
is the mass of the car after the load of gravel is dropped
v2 is the final speed of the car
Solving for v2, we find

Answer:
121Sb=57.2%
123Sb=42.8%
Explanation:
We are given that
Atomic mass of 121Sb=120.904 amu
Atomic mass of 123Sb=122.904 amu
Average atomic mass of antimony=121.760 amu
We have to find the percent of each of the isotope.
Let x be the percent of 121Sb and 1-x be the percent of 123Sb.
Using formula of average atomic weight
Average atomic weight=atomic weight of 121Sb
percentage abundance of isotope 121Sb+atomic weight of 123Sb
percentage abundance of isotope 123Sb
Substitute the values






Percentage of 121Sb=
57.2%
Abundance of isotope 123Sb=1-0.572=0.428
Percentage of isotope 123Sb=
42.8%