Answer: 55.52 *10^-6 C= 55.52 μC
Explanation: In order to solve this question we have to take into account the following expressions:
potential energy stired in a capacitor is given by:
U=Q^2/(2*C) where Q and C are the charge and capacitance of the capacitor.
then we have:
Q^2= 2*C*U=
C=εo*A/d where A and d are the area and separation of the parallel plates capacitor
Q^2=2*εo*A*U/d=2*8.85*10^-12*1.9*10^-5*11*10^3/(1.2*10^-3)=
=55.52 *10^-6C
D: a force equal to the force
Answer: How to solve for FX and FY?
to find fx(x, y): keeping y constant, take x derivative; • to find fy(x, y): keeping x constant, take y derivative. f(x1,...,xi−1,xi + h, xi+1,...,xn) − f(x) h . ∂y2 (x, y) ≡ ∂ ∂y ( ∂f ∂y ) ≡ (fy)y ≡ f22. similar notation for functions with > 2 variables.
Explanation:
Answer:
the resulting angular acceleration is 15.65 rad/s²
Explanation:
Given the data in the question;
force generated in the patellar tendon F = 400 N
patellar tendon attaches to the tibia at a 20° angle 3 cm( 0.03 m ) from the axis of rotation at the knee.
so Torque produced by the knee will be;
T = F × d⊥
T = 400 N × 0.03 m × sin( 20° )
T = 400 N × 0.03 m × 0.342
T = 4.104 N.m
Now, we determine the moment of inertia of the knee
I = mk²
given that; the lower leg and foot have a combined mass of 4.2kg and a given radius of gyration of 25 cm ( 0.25 m )
we substitute
I = 4.2 kg × ( 0.25 m )²
I = 4.2 kg × 0.0626 m²
I = 0.2625 kg.m²
So from the relation of Moment of inertia, Torque and angular acceleration;
T = I∝
we make angular acceleration ∝, subject of the formula
∝ = T / I
we substitute
∝ = 4.104 / 0.2625
∝ = 15.65 rad/s²
Therefore, the resulting angular acceleration is 15.65 rad/s²
Answer:
Approximately 
.
Explanation:
Cathode is where reduction takes place and anode is where oxidation takes place. The potential of a electrochemical reaction (
) is equal to
.
There are two half-reactions in this question. 
and 
. Either could be the cathode (while the other acts as the anode.) However, for the reaction to be spontaneous, the value of should be positive.
In this case, is positive only if is the reaction takes place at the cathode. The net reaction would be
.
Its cell potential would be equal to 
.
The maximum amount of electrical energy possible (under standard conditions) is equal to the free energy of this reaction:
,
where
is the number moles of electrons transferred for each mole of the reaction. In this case the value of is 
as in the half-reactions.
is Faraday's Constant (approximately 
.)
.
