Answer:
Speed of the boat, v = 4.31 m/s
Explanation:
Given that,
Height of the bridge, h = 32 m
The model boat is 11 m from the point of impact when the key was released, d = 11 m
Firstly, we will find the time needed for the boat to get in this position using second equation of motion as :

Here, u = 0 and a = g


t = 2.55 seconds
Let v is the speed of the boat. It can be calculated as :


v = 4.31 m/s
So, the speed of the boat is 4.31 m/s. Hence, this is the required solution.
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:Atomic model keep changing because the electrons around the nucleus are not fixed and they keeps rotating or changing their position in valence orbits around the nucleus.
Explanation: