Answer:
The mass of the sand that will fall on the disk to decrease the is 0.3375 kg
Explanation:
Moment before = Moment after
where;
I is moment of inertia = Mr² = 0.3 x (0.3)² = 0.027 kg.m²
substitute this in the above equation;
![m = \frac{ 0.027[3(2 \pi) - 2(2 \pi)]} {0.2^2 * 6\pi } = \frac{ 0.027[6 \pi - 4\pi]} {0.2^2 * 4\pi }\\\\m = 0.3375kg](https://tex.z-dn.net/?f=m%20%3D%20%5Cfrac%7B%200.027%5B3%282%20%5Cpi%29%20%20-%202%282%20%5Cpi%29%5D%7D%20%7B0.2%5E2%20%2A%206%5Cpi%20%7D%20%3D%20%5Cfrac%7B%200.027%5B6%20%5Cpi%20%20-%204%5Cpi%5D%7D%20%7B0.2%5E2%20%2A%204%5Cpi%20%7D%5C%5C%5C%5Cm%20%3D%200.3375kg)
Therefore, the mass of the sand that will fall on the disk to decrease the is 0.3375 kg
Answer:
Armando's weight ,restored force created by the trampoline
a harmonic movement within the trampoline
Explanation:
In a trampoline we have two forces that actuate Armando's weight and the restored force created by the trampoline that depends on the deformation distance of the elastic canvas.
Amando's weight is vertical and directed towards the center of the Earth and has a constant value, this weight is balanced with the elastic force the springboard exerts on Armando in a vertical direction.
In general, when entering the trampoline, a small jump is made, this creates a speed that deforms the canvas until the speed is reduced to zero, at this point the elastic force is greater than the weight and the boy begins to climb, After the boy leaves the canvas he meets only the force of gravity and his speed decreases to zero and begins his fall.
In Summary Armando is in a harmonic movement within the trampoline
Answer:
1. E x 4πr² = ( Q x r³) / ( R³ x ε₀ )
Explanation:
According to the problem, Q is the charge on the non conducting sphere of radius R. Let ρ be the volume charge density of the non conducting sphere.
As shown in the figure, let r be the radius of the sphere inside the bigger non conducting sphere. Hence, the charge on the sphere of radius r is :
Q₁ = ∫ ρ dV
Here dV is the volume element of sphere of radius r.
Q₁ = ρ x 4π x ∫ r² dr
The limit of integration is from 0 to r as r is less than R.
Q₁ = (4π x ρ x r³ )/3
But volume charge density, ρ = 
So, 
Applying Gauss law of electrostatics ;
∫ E ds = Q₁/ε₀
Here E is electric field inside the sphere and ds is surface element of sphere of radius r.
Substitute the value of Q₁ in the above equation. Hence,
E x 4πr² = ( Q x r³) / ( R³ x ε₀ )
Answer:
The angular acceleration of the pencil<em> α = 17 rad·s⁻²</em>
Explanation:
Using Newton's second angular law or torque to find angular acceleration, we get the following expressions:
τ = I α (1)
W r = I α (2)
The weight is that the pencil has is,
sin 10 = r / (L/2)
r = L/2(sin(10))
The shape of the pencil can be approximated to be a cylinder that rotates on one end and therefore its moment of inertia will be:
I = 1/3 M L²
Thus,
mg(L / 2)sin(10) = (1/3 m L²)(α)
α(f) = 3/2(g) / Lsin(10)
α = 3/2(9.8) / 0.150sin(10)
<em> α = 17 rad·s⁻²</em>
Therefore, the angular acceleration of the pencil<em> </em>is<em> 17 rad·s⁻²</em>
