Answer:
Label A: Battery, Label B: Light or Bulb, Label C: Switch
Explanation:
I got it right.
Answer:
x(t) = - 6 cos 2t
Explanation:
Force of spring = - kx
k= spring constant
x= distance traveled by compressing
But force = mass × acceleration
==> Force = m × d²x/dt²
===> md²x/dt² = -kx
==> md²x/dt² + kx=0 ------------------------(1)
Now Again, by Hook's law
Force = -kx
==> 960=-k × 400
==> -k =960 /4 =240 N/m
ignoring -ve sign k= 240 N/m
Put given data in eq (1)
We get
60d²x/dt² + 240x=0
==> d²x/dt² + 4x=0
General solution for this differential eq is;
x(t) = A cos 2t + B sin 2t ------------------------(2)
Now initially
position of mass spring
at time = 0 sec
x (0) = 0 m
initial velocity v= = dx/dt= 6m/s
from (2) we have;
dx/dt= -2Asin 2t +2B cost 2t = v(t) --- (3)
put t =0 and dx/dt = v(0) = -6 we get;
-2A sin 2(0)+2Bcos(0) =-6
==> 2B = -6
B= -3
Putting B = 3 in eq (2) and ignoring first term (because it is not possible to find value of A with given initial conditions) - we get
x(t) = - 6 cos 2t
==>
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)]