The moment of inertia of a point mass about an arbitrary point is given by:
I = mr²
I is the moment of inertia
m is the mass
r is the distance between the arbitrary point and the point mass
The center of mass of the system is located halfway between the 2 inner masses, therefore two masses lie ℓ/2 away from the center and the outer two masses lie 3ℓ/2 away from the center.
The total moment of inertia of the system is the sum of the moments of each mass, i.e.
I = ∑mr²
The moment of inertia of each of the two inner masses is
I = m(ℓ/2)² = mℓ²/4
The moment of inertia of each of the two outer masses is
I = m(3ℓ/2)² = 9mℓ²/4
The total moment of inertia of the system is
I = 2[mℓ²/4]+2[9mℓ²/4]
I = mℓ²/2+9mℓ²/2
I = 10mℓ²/2
I = 5mℓ²
To solve this problem we will apply the concepts related to Ohm's law and Electric Power. By Ohm's law we know that resistance is equivalent to,
Here,
V = Voltage
I = Current
While the power is equivalent to the product between the current and the voltage, thus solving for the current we have,
Applying Ohm's law
Therefore the equivalent resistance of the light string is 
The answer is Decibels. <span />
meter, millimeters, kilometers. liters. kilograms. centimeters etc... look up the rest
Answer:
friction, conduction and induction
Explanation:
Had it in class I had it correct hope this helps.
