Answer:
w = √[g /L (½ r²/L2 + 2/3 ) ]
When the mass of the cylinder changes if its external dimensions do not change the angular velocity DOES NOT CHANGE
Explanation:
We can simulate this system as a physical pendulum, which is a pendulum with a distributed mass, in this case the angular velocity is
w² = mg d / I
In this case, the distance d to the pivot point of half the length (L) of the cylinder, which we consider long and narrow
d = L / 2
The moment of inertia of a cylinder with respect to an axis at the end we can use the parallel axes theorem, it is approximately equal to that of a long bar plus the moment of inertia of the center of mass of the cylinder, this is tabulated
I = ¼ m r2 + ⅓ m L2
I = m (¼ r2 + ⅓ L2)
now let's use the concept of density to calculate the mass of the system
ρ = m / V
m = ρ V
the volume of a cylinder is
V = π r² L
m = ρ π r² L
let's substitute
w² = m g (L / 2) / m (¼ r² + ⅓ L²)
w² = g L / (½ r² + 2/3 L²)
L >> r
w = √[g /L (½ r²/L2 + 2/3 ) ]
When the mass of the cylinder changes if its external dimensions do not change the angular velocity DOES NOT CHANGE
Answer:
Displacement by cyclist is zero.
Explanation:
In the given question bicyclist is travelling in a rectangular track having P , Q and R edges.
The bicyclist starts from P and travel through Q and R and returned to P again.
We need to find its displacement.
We know displacement of a body is its difference between its initial position to final position.
Here in the given question the bicyclist returns to P again.
Therefore, total displacement by bicyclist is zero.
Hence, this is the required solution.
Answer:
CAN YOU PLEASE GIVE ME POINTS???
Explanation:
<h2>
Answer:</h2>
1.8 x 10⁻⁵J
<h2>
Explanation:</h2>
The energy (E) stored in a capacitor of capacitance, C, when a voltage, V, is supplied is given by;
E = 
x C x V² -------------------(i)
Now, from the question;
C = 2.00μF = 2.00 x 10⁻⁶F
V = 18.0V
Substitute these values into equation (i) as follows;
E = x 2.00 x 10⁻⁶ x 18.0
E = 1.8 x 10⁻⁵J
Therefore, the quantity of energy stored in the capacitor is 1.8 x 10⁻⁵J
