A net force is applied on a 100 kg rocket which causes the rocket to acceleration at 10 m/s2. The same net force is applied on a
1 answer:
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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:
Explanation:
The attached image shows the system expressed in the question.
We can define an expression for the system.
The equivalent equation for the system would be
so, the input signal could be expressed in dB terms
(1)
so the output signal could be expressed as.
The gain should be expressed in dB terms and power in dBm terms so
using the (1) equation to find it in terms of Watts
We can solve the problem by using the law of conservation of energy.
Using the ground as reference point, the mechanical energy of the brick when it is at 5 m from the ground is just potential energy (because the brick is initially at rest, so it doesn't have kinetic energy):
when the brick is at h'=3 m from the ground, its mechanical energy is now sum of kinetic energy and potential energy:
where v is the velocity of the brick. Since E is conserved, it must be equal to the initial energy (98.1 J), so we can solve this equation to find v:
Using the ground as reference point, the mechanical energy of the brick when it is at 5 m from the ground is just potential energy (because the brick is initially at rest, so it doesn't have kinetic energy):
when the brick is at h'=3 m from the ground, its mechanical energy is now sum of kinetic energy and potential energy:
where v is the velocity of the brick. Since E is conserved, it must be equal to the initial energy (98.1 J), so we can solve this equation to find v: