Answer:
Explanation:
Given that,
Charge q=-5.90nC
Magnetic field B= -1.28T k
And the magnetic force
F =−( 3.70×10−7N )i+( 7.60×10−7N )j
Let the velocity be V(xi + yj + zk)
Then, the force is given as
Note i×i=j×j×k×k=0
i×j=k. j×i=-k
j×k=i. k×j=-i
k×i=j. i×k=-j
The force in a magnetic field is given as
F= q(v×B)
−( 3.70×10−7N )i+( 7.60×10−7N )j =
q(xi + yj + zk) × -1.28k
−( 3.70×10−7N )i+( 7.60×10−7N )j=
q( -1.28x i×k - 1.28y j×k - 1.28z k×k)
−( 3.70×10−7N )i+( 7.60×10−7N )j=
q( 1.28xj - 1.28y i )
−( 3.70×10−7N )i+( 7.60×10−7N )j=
q( -1.28y i + 1.28x j)
So comparing comparing coefficients
let compare x axis component
-( 3.70×10−7N )i=-1.28qy i
−3.70×10−7N = -1.28qy
y= -3.7×10^-7/-1.28q
y= -3.7×10^-7/-1.28×-5.90×10^-9)
y=-48.99m/s
y≈-49m/s
Let compare y-axisaxis
7.6×10−7N j = 1.28qx j
7.6×10−7N = 1.28qx
x= 7.6×10^-7/1.28q
x= 7.6×10^-7/1.28×-5.90×10^-9
x=-100.64m/s
a. Then, the velocity of the x component is x= -100.64m/s
b. Also, the velocity component of the y axis is y=-49m/s
c. We will compute
V•F
V=-100.64i -49j
F=−( 3.70×10−7 N )i+( 7.60×10−7 N )j
Note
i.j=j.i=0. Also i.i = j.j =1
V•F is
(-100.64i-49j)•−(3.70×10−7N)i+(7.60×10−7 N )j =
3.724×10^-5 - 3.724×10^-5=0
V•F=0
d. Angle between V and F
V•F=|V||F|Cosx
0=|V||F|Cosx
Cosx=0
x= arccos(0)
x=90°
Since the dot product is zero, from vectors , if the dot product of two vectors is zero, then the vectors are perpendicular to each other
is the number moles of electrons transferred for each mole of the reaction. In this case the value of
as in the half-reactions.
is Faraday's Constant (approximately
.)