Answer:
0.087 m
Explanation:
Length of the rod, L = 1.5 m
Let the mass of the rod is m and d is the distance between the pivot point and the centre of mass.
time period, T = 3 s
the formula for the time period of the pendulum is given by
.... (1)
where, I is the moment of inertia of the rod about the pivot point and g is the acceleration due to gravity.
Moment of inertia of the rod about the centre of mass, Ic = mL²/12
By using the parallel axis theorem, the moment of inertia of the rod about the pivot is
I = Ic + md²

Substituting the values in equation (1)


12d² -26.84 d + 2.25 = 0


d = 2.15 m , 0.087 m
d cannot be more than L/2, so the value of d is 0.087 m.
Thus, the distance between the pivot and the centre of mass of the rod is 0.087 m.
Answer:
At 81. 52 Deg C its resistance will be 0.31 Ω.
Explanation:
The resistance of wire =
Where
=Resistance of wire at Temperature T
= Resistivity at temperature T ![=\rho_0 \ [1 \ + \alpha\ (T-T_0\ )]](https://tex.z-dn.net/?f=%3D%5Crho_0%20%5C%20%5B1%20%5C%20%2B%20%5Calpha%5C%20%28T-T_0%5C%20%29%5D)
Where 
l=Length of the wire
& A = Area of cross section of wire
For long and thin wire the resistance & resistivity relation will be as follows

![\frac{0.25}{0.31}=\frac{1}{[1+\alpha(T-20)]}](https://tex.z-dn.net/?f=%5Cfrac%7B0.25%7D%7B0.31%7D%3D%5Cfrac%7B1%7D%7B%5B1%2B%5Calpha%28T-20%29%5D%7D)



T = 81.52 Deg C
Hello!
First one we can use that PE=mgh so we have
4.37*10^5J/(9.12*10^3kg*9.80m/s^2)= 4.89m
Second one we can use Newton’s Second Law
F=ma and in this case F=mg so we have
g= 3.28*10^-2N/6*10^-3kg = 5.47m/s^2
Hope this helps. Any questions please ask. Thank you.
The answer is Decibels. <span />