The net force acting on a Cessna 172 airplane has a magnitude of 1900 N and points in the positive x direction. If the plane has
1 answer:
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Answer:
F= σ² L² /2ε₀
F = (L² ε₀/4π) ΔV² / r⁴
Explanation:
a) For this exercise we can use Coulomb's law
F = - k Q² / r²
where the negative sign indicates that the force is attractive and the value of the charge is equal to the two plates
Capacitance is defined by
C = Q / ΔV
Q = C ΔV
also the capacitance for a parallel plate capacitor is related to its shape
C = ε₀ A / r
we substitute
Q = ε₀ A ΔV / r
we substitute in the force equation
F = k (ε₀ A ΔV / r)² / r²
k = 1 / 4πε₀
F = ε₀ /4π L² ΔV² / r⁴4
F = L² ΔV² ε₀/ (4π r⁴)
F = (L² ε₀/4π) ΔV² / r⁴
b) Another way to solve the exercise is to use the relationship between the force and the electric field
F = q E
where we can calculate the field created by a plane using Gaussian law, where we use a cylinder with a base parallel to the plate as the Gaussian surface
Ф = ∫E .dA = / ε₀
the plate have two side
2E A = q_{int} / ε₀
E = σ / 2ε₀
σ = q_{int} / A
substituting in force
F = q σ / 2ε₀
the charge total on the other plate is
q = σ A
q = σ L²
F= σ² L² /2ε₀
Answer:
66.98 db
Explanation:
We know that
L_T= Total signal level in db
n= number of sources
L_S= signal level from signal source.
= 66.98 db
Answer:
A)
B)
C)
Explanation:
This question is incomplete. The full question was:
<em>A skateboarder with mass ms = 54 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then jumps on his skateboard and descends down the ramp. His speed at the bottom of the ramp is vf = 6.2 m/s. </em>
<em>Part (a) Write an expression for the work, Wf, done by the friction force between the ramp and the skateboarder in terms of the variables given in the problem statement. </em>
<em>Part (b) The ramp makes an angle θ with the ground, where θ = 30°. Write an expression for the magnitude of the friction force, fr, between the ramp and the skateboarder. </em>
<em>Part (c) When the skateboarder reaches the bottom of the ramp, he continues moving with the speed vf onto a flat surface covered with grass. The friction between the grass and the skateboarder brings him to a complete stop after 5.00 m. Calculate the magnitude of the friction force, Fgrass in newtons, between the skateboarder and the grass.</em>
For part A), we make a balance of energy to calculate the work done by the friction force:
For part B), we use our previous value for the work:
Solving for friction force:
For part C), we first calculate the acceleration by kinematics and then calculate the module of friction force by dynamics:
Solving for a:
Now, by dynamics: