When a voltage difference is applied to a piece of metal wire, a 5.0 mA current flows through it. If this metal wire is now repl
1 answer:
You might be interested in
Answer:
a) -1.325 μC
b) 4.17μC
Explanation:
First, you need to know that charge is conserved. So, the adition of the charges, as there is no lost in charge, should always be the same. Also, after the wire is removed, both spheres will have the same charge, trying to find equilibrium. In summary:
We know both q1f and q2f must be positive, because the negative charge at the beginning was the the smaller.
The electrostatic force is equal to:
K is the Coulomb constant, equal to 9*10^9 Nm^2/C^2
Now, we are told that the electrostatic force after the wire is equal to 0.0443 N:
Originally, the force is negative because it was an attraction force, therefore, its direction was opposite to the direction of the repulsive force after the wire:
Solving the quadratic equation:
for this values q_1 wil be:
So as you can see, the negative charge will always be -1.325 μC and the positive 4.17μC
Answer:
Because the light reflects multiple times until it gets to the Cassegrain focus.
Explanation:
The Cassegrain design can be seen in a reflecting telescope. In this type of design the light is collected by a concave mirror, and then intercepted by a secondary convex mirror, and sends it down to a central opening in the primary mirror (concave mirror), in which a detector is placed (Cassegrain focus)
Since, the light is reflected many times due to Cassegrain design, that leads to shorter telescopes.
