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levacccp [35]
3 years ago
15

The graph shows the amplitude of a passing wave over time in seconds (s) What is the approximate frequency of the wave shown? A.

6 Hz B. 0.2 0.4 Hz
Physics
1 answer:
Lera25 [3.4K]3 years ago
4 0

Answer:

0.2 I took the test

Explanation:

I took the test

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Answer:

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4 years ago
Establishing a potential difference The deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 10
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Answer:

t = 23.9nS

Explanation:

given :

Area A= 10 cm by 2 cm => 2 x 10^-2m x 10 x 10^-2m

distance d= 1mm=> 0.001

resistor R= 975 ohm

Capacitance can be calculated through the following formula,

C = (ε0  x A )/d

C = (8.85 x 10^-12 x (2 x 10^-2 x 10 x 10^-2))/0.001

C = 17.7 x 10^-12    (pico 'p' = 10^-12)

C = 17.7pF

the voltage between two plates is related to time, There we use the following formula of the final voltage

Vc = Vx (1-e^-(t/CR))  

75 = 100 x (1-e^-(t/CR))

75/100 = (1-e^-(t/CR))

.75 = (1-e^-(t/CR))

.75 -1 = -e^-(t/CR)

-0.25 = -e^-(t/CR)  --->(cancelling out the negative sign)

e^-(t/CR) = 0.25

in order to remove the exponent, take logs on both sides  

-t/CR = ln (0.25)

t/CR = -ln(0.25)

t = -CR x ln (0.25)

t = -(17.7 x 10^-12 x 975) x (-1.38629)

t = 23.9 x 10^{-9  

t = 23.9ns

Thus, it took 23.9ns  for the potential difference between the deflection plates to reach 75 volts

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