Do the math: How many seconds would it take an echo sounder’s ping to make the trip from a ship to the Challenger Deep (10,994 m
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The formula for the period of wave is: wave period is equals to 1 over the frequency.
To get the value of period of wave you need to divide 1 by 200 Hz. However, beforehand, you have to convert 200 Hz to cycles per second. So that would be, 200 cyles per second or 200/s.
By then, you can start the computation by dividing 1 by 200/s. Since 200/s is in fractional form, you have to find its reciprocal form and multiply it to one which would give you 1 (one) second over 200. This would then lead us to the value 0.005 seconds as the wave period.
wave period= 1/200 Hz
Convert Hz to cycles per second first
200 Hz x 1/s= 200/second
Make 200/second as your divisor, so:
wave period= 1/ 200/s
get the reciprocal form of 200/s which is s/200
then you can start the actual computation:
wave period= 1 x s divided by 200
this would give us an answer of 0.005 s.
To get the value of period of wave you need to divide 1 by 200 Hz. However, beforehand, you have to convert 200 Hz to cycles per second. So that would be, 200 cyles per second or 200/s.
By then, you can start the computation by dividing 1 by 200/s. Since 200/s is in fractional form, you have to find its reciprocal form and multiply it to one which would give you 1 (one) second over 200. This would then lead us to the value 0.005 seconds as the wave period.
wave period= 1/200 Hz
Convert Hz to cycles per second first
200 Hz x 1/s= 200/second
Make 200/second as your divisor, so:
wave period= 1/ 200/s
get the reciprocal form of 200/s which is s/200
then you can start the actual computation:
wave period= 1 x s divided by 200
this would give us an answer of 0.005 s.
Answer:
a) the required time is 0.6283 μs
b) the inductor current is 0.5 mA
Explanation:
Given the data in the question;
The capacitor voltage has its maximum value of 25 V at t = 0
i.e V = V₀ = 25 V
we determine the angular velocity;
ω = 1 / √( LC )
ω = 1 / √( ( 20 × 10⁻³ H ) × ( 8.0 × 10⁻¹² F) )
ω = 1 / √( 1.6 × 10⁻¹³ )
ω = 1 / 0.0000004
ω = 2.5 × 10⁶ s⁻¹
a) How much time does it take until the capacitor is fully discharged for the first time?
V = V₀sin( ωt )
we substitute
25V = 25V × sin( 2.5 × 10⁶ s⁻¹ × t )
25V = 25V × sin( 2.5 × 10⁶ s⁻¹ × t )
divide both sides by 25 V
sin( 2.5 × 10⁶ × t ) = 1
( 2.5 × 10⁶ × t ) = π/2
t = 1.570796 / (2.5 × 10⁶)
t = 0.6283 × 10⁻⁶ s
t = 0.6283 μs
Therefore, the required time is 0.6283 μs
b) What is the inductor current at that time?
(t) = V₀√(C/L) sin(ωt)
{ sin(ωt) = 1 )
(t) = V₀√(C/L)
we substitute
(t) = 25V × √( ( 8.0 × 10⁻¹² F ) / ( 20 × 10⁻³ H ) )
(t) = 25 × 0.00002
(t) = 0.0005 A
(t) = 0.5 mA
Therefore, the inductor current is 0.5 mA