Answer:
The Answer is A Hope I helped you :D Have a Great Day!
Explanation:
Answer:False
Explanation:
Work is being done on a body when it causes displacement of body on the application of force

When we pull the door by a force it causes zero displacements of the door. So we can say that work done on it is zero.
Thus the above-given statement is false
Answer:
8.0 N
Explanation:
Force: This can be defined as the mass of a body and its acceleration. The S.I unit of Force is Newton (N).
Mathematically, Fore is expressed as
F = ma ........................... equation 1
Where F = force, m = mass, a = acceleration.
and
I = mΔv
Δv = I/m ............................ Equation 2
Where I = impulse, m = mass, Δv = change in velocity
Given: I = 6.0 Newton-seconds, m = 0.1 kilogram.
Substituting into equation 2
Δv = 6.0/0.1
Δv = 60 m/s.
But
a = Δv/t
where t = time = 0.75 seconds.
a = 60/0.75
a = 80 m/s²
Substitute the values of a and m into equation 1.
F = 0.1(80)
F = 8.0 N.
Thus the average force produced = 8.0 N
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)]