Answer:
The spring constant = 104.82 N/m
The angular velocity of the bar when θ = 32° is 1.70 rad/s
Explanation:
From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:


Also;

Thus;

where;
= deflection in the spring
k = spring constant
b = remaining length in the rod
m = mass of the slender bar
g = acceleration due to gravity


Thus; the spring constant = 104.82 N/m
b
The angular velocity can be calculated by also using the conservation of energy;






Thus, the angular velocity of the bar when θ = 32° is 1.70 rad/s
the electric conductivity of gold is very high
Explanation:
It is given that,
Magnitude of charge, 
It moves in northeast direction with a speed of 5 m/s, 25 degrees East of a magnetic field.
Magnetic field, 
Velocity, 
![v=[(4.53)i+(2.11)j]\ m/s](https://tex.z-dn.net/?f=v%3D%5B%284.53%29i%2B%282.11%29j%5D%5C%20m%2Fs)
We need to find the magnitude of force on the charge. Magnetic force is given by :

![F=15\times 10^{-6}[(4.53i+2.11j)\times 0.08\ j]](https://tex.z-dn.net/?f=F%3D15%5Ctimes%2010%5E%7B-6%7D%5B%284.53i%2B2.11j%29%5Ctimes%200.08%5C%20j%5D)
<em>Since</em>, 
![F=15\times 10^{-6}[(4.53i)\times (0.08)\ j]](https://tex.z-dn.net/?f=F%3D15%5Ctimes%2010%5E%7B-6%7D%5B%284.53i%29%5Ctimes%20%280.08%29%5C%20j%5D)


So, the force acting on the charge is
and is moving in positive z axis. Hence, this is the required solution.