The biggest reason that some fuels are considered nonrenewable
1 answer:
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Answer:
a
The radial acceleration is
b
The horizontal Tension is
The vertical Tension is
Explanation:
The diagram illustrating this is shown on the first uploaded
From the question we are told that
The length of the string is
The mass of the bob is
The angle made by the string is
The centripetal force acting on the bob is mathematically represented as
Now From the diagram we see that this force is equivalent to
where T is the tension on the rope and v is the linear velocity
So
Now the downward normal force acting on the bob is mathematically represented as
So
=>
=>
The centripetal acceleration which the same as the radial acceleration of the bob is mathematically represented as
=>
substituting values
The horizontal component is mathematically represented as
substituting value
The vertical component of tension is
substituting value
The vector representation of the T in term is of the tension on the horizontal and the tension on the vertical is
substituting value
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)]