Answer:
b) 472HZ, 408HZ
Explanation:
To find the frequencies perceived when the bus approaches and the train departs, you use the Doppler's effect formula for both cases:
fo: frequency of the source = 440Hz
vs: speed of sound = 343m/s
vo: speed of the observer = 0m/s (at rest)
v: sped of the train
f: frequency perceived when the train leaves us.
f': frequency when the train is getTing closer.
Thus, by doing f and f' the subjects of the formulas and replacing the values of v, vo, vs and fo you obtain:
hence, the frequencies for before and after tha train has past are
b) 472HZ, 408HZ
Answer:
same 0.81m
Explanation:
in this problem if we assume there no resistance of any sort. and we apply the energy conservation
change in Potential energy = change in kinetic energy
mgh = 0.5mv^2
gh = 0.5v^2
the above relation suggests that the speed at the bottom is only depending on the height it is released from not on the shape, mass or radius.
so at the bottom
put h = 0.81m
9.81 * 0.81 * 2 = v^2
v=3.99 m/s
both CYLINDER and SPHERE will have same velocity at the bottom if released from the same height irrespective of shape and size
(a) The time for the capacitor to loose half its charge is 2.2 ms.
(b) The time for the capacitor to loose half its energy is 1.59 ms.
<h3>
Time taken to loose half of its charge</h3>
q(t) = q₀e-^(t/RC)
q(t)/q₀ = e-^(t/RC)
0.5q₀/q₀ = e-^(t/RC)
0.5 = e-^(t/RC)
1/2 = e-^(t/RC)
t/RC = ln(2)
t = RC x ln(2)
t = (12 x 10⁻⁶ x 265) x ln(2)
t = 2.2 x 10⁻³ s
t = 2.2 ms
<h3>
Time taken to loose half of its stored energy</h3>
U(t) = Ue-^(t/RC)
U = ¹/₂Q²/C
(Ue-^(t/RC))²/2C = Q₀²/2Ce
e^(2t/RC) = e
2t/RC = 1
t = RC/2
t = (265 x 12 x 10⁻⁶)/2
t = 1.59 x 10⁻³ s
t = 1.59 ms
Thus, the time for the capacitor to loose half its charge is 2.2 ms and the time for the capacitor to loose half its energy is 1.59 ms.
Learn more about energy stored in capacitor here: brainly.com/question/14811408
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Answer:
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