Explanation:
Given data
Inductance L=12*10^-³H
Capacitance C= 3.5*10^-6F
Resistance R= 3.3 Ohms
Voltage V=115v
Capacitive reactance Xc=?
inductive reactance Xl=?
Impedance Z=?
Phase angle =?
A. Resonance frequency
In RLC circuit resonance occurs when capacitive reactance equals inductive reactance
f=1/2pi √ LC
f=1/2*3.142 √ 12*10^-³*3.5*10^-6
f=1/6.284*0.0002
f=1/0.00125
f=800HZ
B. Find Irms at resonance.
Irms=R/V
Irms=3.3/115
Irms=0.028amp
Find the capacitive reactance XC in Ohms
Xc=1/2pi*f*C
Xc=1/2*3.142*800*3.5*10^-6
Xc=1/0.0176
Xc=56.8 ohms
To find the inductive reactance
Xl=2pifL
Xl=2*3.142*800*12*10^-3
Xl=60.3ohms
d) Find the impedance Z.
Z=√R²+(Xl-Xc)²
Z=√3.3²+(60.3-56.8)²
Z=√10.89+12.25
Z=√23.14
Z=4.8ohms
Phase angle =
Tan phi=Xc/R=56.8/3.3
Tan phi=17.2
Phi=tan-1 17.2
Phi= 1.51°
Answer: (a) t = 5.44 sec
(b) vf = 53.31 m/s
(c) s = 5.0m
Explanation: from the question, given data
the Height of the tower, h = 145m
from question
(a)
the initial velocity, v₁ = 0 m/s
s = v₁t + 1/2 gt²
-145 m = 0(t) + 1/2 (-9.8t²)
t² = 145/4.9
t² = 29.59
t = 5.44 sec
(b)
the speed of the sphere at the bottom of the tower is
vf² = vi² +2as
vf² = 0 + 2(-9.8 × -145)
vf² = 2842
vf = 53.31 m/s
(c)
when caught, the sphere experiences a deceleration of;
a = -29.0g
the time it would take to decelerate becomes;
vf = vi + at
0 = (53.31) + (-29 ×9.8)t
where t = 53.31 / 284.2
t = 0.1876 sec
∴ the distance travelled during the deceleration becomes;
vf² = vi² + 2as
s = (vf² - vi²) / 2a
s = (0 - 53.31²) / 2×-29×9.8
s = -2841.9561 / -568.4
s = 4.99 ≈ 5.0m
i hope this helps, cheers
Answer:
The answer to this is falling all the way through the Earth is impossible, since its core is molten. ... As you approached the center of the earth the pull of gravity would decline and eventually (at the center) cease, but inertia would keep you going.
Explanation:
your welcome
Answer:
The answer is B).
Explanation:
The correct way to write this sentence is: After I woke up this morning, I got me a tall glass of orange juice.
Explanation:
Unclear question. The clear rendering reads;
"Into a U-tube containing mercury, pour on the other side sulfuric acid of density 1.84 and on the other side alcohol of density 0.8 so that the levels are in the same horizontal plane. The height of the acid above the mercury being 24 cm. What is the height of the bar and what variation of the level of the acid, when the mercury density is 13.6?
