All three windows are the same size.
A has 10 complete waves visible through the window. B has 3, and C has 4.
So A must have the smallest wavelengths.
Given that the block have two applied masses 250 g at East and 100 g at South. In order to make a situation in which block moves towards point A, we have to apply minimum number of masses to the blocks. In order to prevent block moving toward East, we have to apply a mass at West, equal to the magnitude of mass at East but opposite in direction. Therefore, mass of 250 g at West is the required additional mass that has to be added. There is already 100 g of mass acting at South, that will attract block towards South or point A. No need to add further mass in North-South direction.
Answer:
B) t = 1.83 [s]
A) y = 16.51 [m]
Explanation:
To solve this problem we must use the following equation of kinematics.

where:
Vf = final velocity = 0
Vo = initial velocity = 18 [m/s]
g = gravity acceleration = 9.81 [m/s²]
t = time [s]
Note: the negative sign in the above equation means that the acceleration of gravity is acting in the opposite direction to the motion.
A) The maximum height is reached when the final velocity of the ball is zero.
0 = 18 - (9.81*t)
9.81*t = 18
t = 18/9.81
t = 1.83 [s], we found the answer for B.
Now using the following equation.

where:
y = elevation [m]
Yo = initial elevation = 0
y = 18*(1.83) - 0.5*9.81*(1.83)²
y = 16.51 [m]
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)]