Answer:
Option D: 21.8 degrees
Explanation:
In a parallel RL circuit, the current in the resistor R and that in the inductor L are separated among themselves 90 degrees as illustrated in the attached image. In the image the current in the resistor is represented in orange, that of the inductor in blue, and the total current (vector addition of the previous two) is represented in red, forming a certain angle (theta) with respect to the current in the resistor. The output voltage is the same as the input voltage as measured over the resistor R.
Therefore, the phase angle that separated output voltage and total current can be obtained using the fact that tan(phase angle) = 
, therefore the angle is the arctangent of 4/10:
degrees.
Answer:
v=30 m/s
Explanation:
h - height
g - acceleration due to gravity=10
t - time
v- velocity
45 = 5t²
t² = 9
t=3 seconds
v=g×t
v=10×3
v=30 m/s
Sound at 70 dB is 70 dB louder than the human reference level. That's 10⁷ times as much as the reference sound power.
Sound at 73 dB is 73 dB louder than the human reference level. That's 10⁷.³ or 2 x 10⁷ times as much as the reference sound power.
Sound at 80 dB is 80 dB louder than the human reference level. That's 10⁸ or 10 x 10⁷ times as much as the reference sound power.
Now we can adumup:
Intensity of all 3 sources = (10⁷) + (2 x 10⁷) + (10 x 10⁷)
Intensity = (13 x 10⁷) times the sound power reference intensity.
Intensity in dB = 10 log (13 x 10⁷) = 10 (7 + log(13)
Intensity = 70 + 10 log(13)
Intensity = 70 + 10 (1.114)
Intensity = 70 + 11.14
Intensity = <em>81.14 dB</em>
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Looking at the questioner's profile, I seriously wonder whether I'll ever get a comment in return from this creature, and how I'll ever find out if my solution is correct. For that matter, I'm also seriously questioning how and whether my solution will ever be used for anything.
Answer:
2.5
Explanation:
The capacitance of a parallel-plate capacitor filled with dielectric is given by
where
k is the dielectric constant
is the capacitance of the capacitor without dielectric
In this problem,
is the capacitance of the capacitor in air
is the capacitance with the dielectric inserted
Solving the equation for k, we find
