Answer:
<em>There will be an increase in potential difference.</em>
Explanation:
As we know that the potential difference depends upon the capacitance.
ΔV = Q/C
When battery is disconnected the charge remains constant on the plates but the capacitance decreases. As the capacitance has an inverse relation with the potential difference, there will be an increase in it.
In addition to that the potential difference can also be defined as the product of field and distance between the plates. As the charge is constant so the field is constant. Upon increasing the separation between the plates the potential difference will also increased.
Answer:
I attached an image that should help.
Explanation:
Check it out.
I'm stuck on the same question, as well :(
Answer:
A.B = -38
Explanation:
A = 2i + 9j and B = -i - 4j.
So, A.B = (2i + 9j).(-i - 4j)
= 2i.(-i) + 2i.(-4j) + 9j.(-i) + 9j.(-4j)
= -2i.i - 8i.j - 9j.i - 36j.j
since i.i = 1, j.j = 1, i.j = 0 and j.i = 0, we have
A.B = -2(1) - 8(0) - 9(0) - 36(1)
A.B = -2 - 0 - 0 - 36
A.B = -38
Answer:
If the acceleration is constant, the movements equations are:
a(t) = A.
for the velocity we can integrate over time:
v(t) = A*t + v0
where v0 is a constant of integration (the initial velocity), for the distance traveled between t = 0 units and t = 10 units, we can solve the integral:
Where to obtain the actual distance you can replace the constant acceleration A and the initial velocity v0.
