Find fq, the vertical force that pier q exerts on the right end of the bridge.
1 answer:
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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.
Answer:
<em>The motorboat ends up 7.41 meters to the west of the initial position </em>
Explanation:
<u>Accelerated Motion </u>
The accelerated motion describes a situation where an object changes its velocity over time. If the acceleration is constant, then these formulas apply:
The problem provides the conditions of the motorboat's motion. The initial velocity is 6.5 m/s west. The final velocity is 1.5 m/s west, and the acceleration is to the east. Since all the movement takes place in one dimension, we can ignore the vectorial notation and work with the signs of the variables, according to a defined positive direction. We'll follow the rule that all the directional magnitudes are positive to the east and negative to the west. Rewriting the formulas:
Solving the first one for t
We have
Using these values
We now compute x
The motorboat ends up 7.41 meters to the west of the initial position
Answer:
angular frequency of the table must be same as the frequency of the projection of the gum on the wall
Explanation:
Since we know that the projection on the wall is the vertical component of the position of the gum on the rotating table
So here we will say
so the angle made by the radius vector depends on the angular frequency of the disc by which it is rotating
So we can say
so here we can say
so here we can say that
angular frequency of the table must be same as the frequency of the projection of the gum on the wall