Answer: 4.98 m/s
Explanation:
You solve these kinetic energy, potential energy problems by using the fact P.E.+ K.E. = a constant as long as friction is ignored.
PEi = 0 in this case
KEi = ½mVi² = PEf+KEf = mghf + ½mVf²
½1210*8.31² = 1210*9.8*2.26 + ½1210*Vf²
½1210*Vf² = ½1210*8.31² - 1210*9.8*2.26
Vf² = 8.31² - 2*9.8*2.26 = 4.98² so Vf = 4.98m/s
Answer:
Four charges of equal magnitude sitting at the vertices of a square
Explanation:
We can arrive at such a situation by thinking of a simple example first, a configuration of two charges. The force acting on the middle point of a straight line joining the two points(charges) will be zero. That is, the net Electric field will be zero as they cancel out being equal in magnitude and opposite in direction.
Now, we can extend this idea to a square having charge q at each vertex. If we put 'p' at the geometric center, we can see that the Electric fields along the diagonals cancel out due to the charges at the diagonally opposite vertices(refer to the figure attached). Actually, the only requirement is that the diagonally opposite charges are equal.
We can further take this to 3 dimensions. Consider a cube having charges of equal magnitude at each vertex. In this case, the point 'p' will yet again be the geometric center as the Electric field due to the diagonally opposite charges will cancel out.
<span>I believe the answer is 0.29.</span>
In order to make things easier to describe and explain, let's call
the resistance of each bulb 'R', and the battery voltage 'V'.
a). In series, the total resistance is 3R.
In parallel, the total resistance is R/3.
Changing from series to parallel, the total resistance of the circuit
decreases to 1/9 of its original value.
b). In series, the total current is V / (3R) .
In parallel, the total current is 3V / R .
Changing from series to parallel, the total current in the circuit
increases to 9 times its original value.
c). In series, the power dissipated by the circuit is
(V) · V/3R = V² / 3R .
In parallel, the power dissipated by the circuit is
(V) · 3V/R = 3V² / R .
Changing from series to parallel, the power dissipated by
the circuit (also the power delivered by the battery) increases
to 9 times its original value.