Can someone please help me no one ever responds to my question and I need the correct answer. I believe the answer is C but i'm
2 answers:
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Answer:
- quality factor (Q) = 69.99
- inductor = 1.591 x 10⁻⁴ H
- capacitor = 3.248 x 10⁻¹⁰ F
Explanation:
Given;
resonance frequency (F₀) = 700 kHz
resistor, R = 10 Ohm
bandwidth (BW) = 10 kHz
bandwidth (BW)
make L (inductor) the subject of the formula
make C (capacitor) the subject of the formula
quality factor (Q)
quality factor (Q) = 69.99
Answer:
k = 11,564 N / m, w = 6.06 rad / s
Explanation:
In this exercise we have a horizontal bar and a vertical spring not stretched, the bar is released, which due to the force of gravity begins to descend, in the position of Tea = 46º it is in equilibrium;
let's apply the equilibrium condition at this point
Axis y
W_{y} - Fr = 0
Fr = k y
let's use trigonometry for the weight, we assume that the angle is measured with respect to the horizontal
sin 46 = / W
W_{y} = W sin 46
we substitute
mg sin 46 = k y
k = mg / y sin 46
If the length of the bar is L
sin 46 = y / L
y = L sin46
we substitute
k = mg / L sin 46 sin 46
k = mg / L
for an explicit calculation the length of the bar must be known, for example L = 1 m
k = 1.18 9.8 / 1
k = 11,564 N / m
With this value we look for the angular velocity for the point tea = 30º
let's use the conservation of mechanical energy
starting point, higher
Em₀ = U = mgy
end point. Point at 30º
= K -Ke = ½ I w² - ½ k y²
em₀ = Em_{f}
mgy = ½ I w² - ½ k y²
w = √ (mgy + ½ ky²) 2 / I
the height by 30º
sin 30 = y / L
y = L sin 30
y = 0.5 m
the moment of inertia of a bar that rotates at one end is
I = ⅓ mL 2
I = ½ 1.18 12
I = 0.3933 kg m²
let's calculate
w = Ra (1.18 9.8 0.5 + ½ 11,564 0.5 2) 2 / 0.3933)
w = 6.06 rad / s
Answer:
in the direction of the first body
Explanation:
Given:
- mass of the first ball,
- velocity of the first ball,
- mass of the second ball,
- velocity of the second ball,
<u>Now for the head-on inelastic collision:</u>
(since the bodies combine after an inelastic collision.)
in the direction of the first body. (Since the net result is positive as assumed in the equation for the first body)