Answer:
E = 4.72 * 10⁻⁶ Nm²
Explanation:
Parameters given:
Outer radius, R = 3.70cm = 0.037m
Inner radius, r = 3.15cm = 0.0315m
Permittivity of free space, ε₀ = 8.85 * 10⁻¹² C/Nm²
Charge density: 1.22 * 10⁻³ C/m³
The question requires that we solve using Gauss law which states that the net electric field through a closed surface is proportional to the enclosed electric charge.
Hence,
E = Q/Aε₀
Charge Q is given as
Q = ρπ(R² ⁻ r²)L
A = 2π(R - r)L
E = [ρπ(R² ⁻ r²)L]/[2π(R - r)ε₀L]
Using difference of two squares,
(R² ⁻ r²) = (R + r)(R - r)
E =[ρ(R + r)]/(2ε₀)
E = [1.22 * 10⁻³ *(0.0370 + 0.0315)]/(2 * 8.85 * 10⁻¹²)
E = 4.72 * 10⁻⁶ Nm²
Answer:
0.83 ω
Explanation:
mass of flywheel, m = M
initial angular velocity of the flywheel, ω = ωo
mass of another flywheel, m' = M/5
radius of both the flywheels = R
let the final angular velocity of the system is ω'
Moment of inertia of the first flywheel , I = 0.5 MR²
Moment of inertia of the second flywheel, I' = 0.5 x M/5 x R² = 0.1 MR²
use the conservation of angular momentum as no external torque is applied on the system.
I x ω = ( I + I') x ω'
0.5 x MR² x ωo = (0.5 MR² + 0.1 MR²) x ω'
0.5 x MR² x ωo = 0.6 MR² x ω'
ω' = 0.83 ω
Thus, the final angular velocity of the system of flywheels is 0.83 ω.
If the rock is just sitting there and you want to SLIDE it, then you have to push it with a force of at least
(251 kg) x (9.8 m/s²) x (μ) =
(2,459 Newtons) x (the coefficient of static friction on that surface)
Answer:
Explanation:
Weight is actually a force. A force can change depending on its location. A mass remains constant no matter where it is.
A)
F = m * a
m = 250 kg
a = 9.81 m/s^2
F = 250 * 9.81 = 2452.5 N
B)
The acceleration due to gravity on the moon is roughly 1/6 what it is on earth. You can check its value in your notes.
a = 9.81 + (1/6) = 1.635
m = 250
F = 250 * 1.635
F = 408.75
C)
The mass is the same anywhere in the universe.
250 kg
