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)]
Answer:
sweeps out equal areas in equal times.
Explanation:
As we know that there is no torque due to Sun on the planets revolving about the sun
so we will have

now we have

now we also know that

so rate of change in area is given as

so we will have


since angular momentum and mass is constant here so
all planets sweeps out equal areas in equal times.
Answer:5.075N
Explanation:
Mass=0.145kg
Acceleration=35m/s^2
Force=mass x acceleration
Force=0.145 x 35
Force=5.075N
This is the upthrust on an object which is placed inside a fluid
This force act upwards and always push upwards
so the correct answer is given as
D. A force within a fluid that pushes upward
this force is always due to pressure difference at two levels of
at lower level since pressure is more that is why the force is upwards and this upthrust is known as Buoyancy