Explanation:
Given that,
Two resistors of resistance 6 ohm and 3 ohm are connected in series and then in parallel.
For series combination,
For parallel combination,
When 6 ohm and 3 ohm are in series,
When 6 ohm and 3 ohm are in paralle,
So, the equivalent resistance in series combination is 9 ohms and in parallel combination it is 2 ohms.
Answer:
a). 1.218 m/s
b). R=2.8
Explanation:
Momentum of the motion the first part of the motion have a momentum that is:
The final momentum is the motion before the action so:
a).
b).
kinetic energy
Kinetic energy after
Kinetic energy before
Ratio =
Answer:
E = k Q / [d(d+L)]
Explanation:
As the charge distribution is continuous we must use integrals to solve the problem, using the equation of the elective field
E = k ∫ dq/ r² r^
"k" is the Coulomb constant 8.9875 10 9 N / m2 C2, "r" is the distance from the load to the calculation point, "dq" is the charge element and "r^" is a unit ventor from the load element to the point.
Suppose the rod is along the x-axis, let's look for the charge density per unit length, which is constant
λ = Q / L
If we derive from the length we have
λ = dq/dx ⇒ dq = L dx
We have the variation of the cgarge per unit length, now let's calculate the magnitude of the electric field produced by this small segment of charge
dE = k dq / x²2
dE = k λ dx / x²
Let us write the integral limits, the lower is the distance from the point to the nearest end of the rod "d" and the upper is this value plus the length of the rod "del" since with these limits we have all the chosen charge consider
E = k
We take out the constant magnitudes and perform the integral
E = k λ (-1/x)
Evaluating
E = k λ [ 1/d - 1/ (d+L)]
Using λ = Q/L
E = k Q/L [ 1/d - 1/ (d+L)]
let's use a bit of arithmetic to simplify the expression
[ 1/d - 1/ (d+L)] = L /[d(d+L)]
The final result is
E = k Q / [d(d+L)]