Explanation:
Using Ohm's Law and a bit of substitution, we can use voltage divided by current to solve for resistance. Doing that, we'll get 6 Ohm.
A sanding disk with rotational inertia 1.2 × 10−3 kg ⋅ m2 is attached to an electric drill whose motor delivers a torque of magn
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Answer:
61.85 ohm
Explanation:
L = 12 m H = 12 x 10^-3 H, C = 15 x 10^-6 F, Vrms = 110 V, R = 45 ohm
Let ω0 be the resonant frequency.
ω0 = 2357 rad/s
ω = 2 x 2357 = 4714 rad/s
XL = ω L = 4714 x 12 x 10^-3 = 56.57 ohm
Xc = 1 / ω C = 1 / (4714 x 15 x 10^-6) = 14.14 ohm
Impedance, Z =
Z = \sqrt{45^{2}+\left ( 56.57-14.14 )^{2}} = 61.85 ohm
Thus, the impedance at double the resonant frequency is 61.85 ohm.
Consult the attached free body diagram.
By Newton's second law, the net force on the crate acting parallel to the surface is
∑ F[para] = (370 N) cos(-20°) - f = 0
(this is the x-component of the resultant force)
where
• (370 N) cos(-20°) = magnitude of the horizontal component of the pushing force
• f = magnitude of kinetic friction
The crate is moving at a constant speed and thus not accelerating, so the crate is in equilibrium.
Solve for f :
f = (370 N) cos(-20°) ≈ 347.686 N
The net force acting perpendicular to the surface is
∑ F[perp] = n - 1480 N - (370 N) sin(-20°) = 0
(this is the y-component of the resultant force)
where
• n = magnitude of normal force
• 1480 N = weight of the crate
• (370 N) sin(-20°) = magnitude of the vertical component of push
The crate doesn't move up or down, so it's also in equilibrium in this direction.
Solve for n :
n = 1480 N + (370 N) sin(-20°) ≈ 1606.55 N ≈ 1610 N
Then the coefficient of kinetic friction is µ such that
f = µn ⇒ µ = f/n ≈ 0.216
