Answer:
Fₓ = 0, = 0 and
<em> = - 3.115 10⁻¹⁵ N</em>
Explanation:
The magnetic force given by the expression
F = q v xB
the bold are vectors, the easiest analytical way to determine this force in solving the determinant
F = 1.6 10⁻¹⁵ [ i( 0-0) + j (0-0) + k^( 5.8 0.60 - 0.81 67) ]
F =i^0 + j^0 - k^ 3.115 10⁻¹⁵ N
Fₓ = 0
= 0
<em> = - 3.115 10⁻¹⁵ N</em>
Answer:
the period of the physical pendulum is 0.498 s
Explanation:
Given the data in the question;
= 0.61 s
we know that, the relationship between T and angular frequency is;
T = 2π/ω ---------- let this be equation 1
Also, the angular frequency of physical pendulum is;
ω = √(mgL / ) ------ let this equation 2
where m is mass of pendulum, L is distance between axis of rotation and the center of gravity of rod and is moment of inertia of rod.
Now, moment of inertia of thin uniform rod D is;
=
mD²
since we were not given the length of the rod but rather the period of the simple pendulum, lets combine this three equations.
we substitute equation 2 into equation 1
we have;
T = 2π/ω OR T = 2π/√(mgL/) OR T = 2π√(
/mgL)
so we can use =
mD² for moment of inertia of the rod
Since center of gravity of the uniform rod lies at the center of rod
so that L = D.
now, substituting these equations, the period becomes;
T = 2π/√(/mgL) OR T =
OR T = 2π√(2D/3g ) ----- equation 3
length of rod D is still unknown, so from equation 1 and 2 ( period of pendulum ),
we have;
ω = 2π/
OR ω
= √(g/D) OR ω
= 2π√( D/g )
so we simple solve for D/g and insert into equation 3
so we have;
T = √(2/3) ×
we substitute in value of
T = √(2/3) × 0.61 s
T = 0.498 s
Therefore, the period of the physical pendulum is 0.498 s