Explanation:
it is almost zero .this is because the distance and the electrostatic force are inversely proportional
Explanation : Explain each characteristic of sound waves.
Intensity : the intensity of the sound wave is understand as the power carry by sound wave per unit area in the direction perpendicular to that area.
Loudness : loudness is the quality of the loud and soft of the sound wave.
Frequency : Human normal hear sound frequency between 20 Hz to kHz.
Pitch : Pitch is the quality of low and high of sound wave . pitch relates to the frequency of the slowest vibration in the sound wave for simple sound.
A pendulum is not a wave.
-- A pendulum doesn't have a 'wavelength'.
-- There's no way to define how many of its "waves" pass a point
every second.
-- Whatever you say is the speed of the pendulum, that speed
can only be true at one or two points in the pendulum's swing,
and it's different everywhere else in the swing.
-- The frequency of a pendulum depends only on the length
of the string from which it hangs.
If you take the given information and try to apply wave motion to it:
Wave speed = (wavelength) x (frequency)
Frequency = (speed) / (wavelength) ,
you would end up with
Frequency = (30 meter/sec) / (0.35 meter) = 85.7 Hz
Have you ever seen anything that could be described as
a pendulum, swinging or even wiggling back and forth
85 times every second ? ! ? That's pretty absurd.
This math is not applicable to the pendulum.
Answer:
Bar graph
Explanation:
each day collects data so a bar graph would work.
Answer:
it will take 36.12 ms to reduce the capacitor's charge to 10 μC
Explanation:
Qi= C×V
then:
Vi = Q/C = 30μ/20μ = 1.5 volts
and:
Vf = Q/C = 10μ/20μ = 0.5 volts
then:
v = v₀e^(–t/τ)
v₀ is the initial voltage on the cap
v is the voltage after time t
R is resistance in ohms,
C is capacitance in farads
t is time in seconds
RC = τ = time constant
τ = 20µ x 1.5k = 30 ms
v = v₀e^(t/τ)
0.5 = 1.5e^(t/30ms)
e^(t/30ms) = 10/3
t/30ms = 1.20397
t = (30ms)(1.20397) = 36.12 ms
Therefore, it will take 36.12 ms to reduce the capacitor's charge to 10 μC.
