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natita
1 month ago
7

How many times could Haley fly bewteen the two flowers in 1 minute (60 seconds)​

Physics
2 answers:
Sav [3.1K]1 month ago
8 0
The number of times Haley could complete the flight between the two flowers hinges on two fundamental parameters. First, the distance separating the two flowers, and second, Haley’s speed during this flight. The provided time frame is 1 minute, so the outcome is entirely dependent on the positioning of these two flowers and Haley’s flying speed. The shorter the distance between them, the more flights she can achieve; conversely, if they are further apart, Haley will take longer to travel between them.
Yuliya22 [3.3K]1 month ago
7 0
The answer is 30 seconds.
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The deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 100 volt potential difference is sudde
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Explanation:

Data provided:

Area A = 10 cm×2 cm = 20×10⁻⁴ m²

Separation distance d between the plates = 1 mm = 1×10⁻³ m

Battery voltage, or emf = 100 V

Resistance = 1025 ohm

Solution:

In an RC circuit, the voltage across the plates varies with time t. At the outset, the voltage matches that of the battery, V₀ = emf = 100V. However, after a certain time t, both the resistance and capacitance alter this, leading to a final voltage V expressed as

V = V_{0}(1-e^{\frac{-t}{RC} } )\\\frac{V}{V_{0} } = 1-e(^{\frac{-t}{RC} }) \\e^{\frac{-t}{RC} } = 1- \frac{V}{V_{0} }

Applying the natural logarithm to both sides,

e^{\frac{-t}{RC} } = 1- \frac{V}{V_{0} } \\\frac{-t}{RC} = ln(1-\frac{V}{V_{0} } )\\t = -RCln(1 - \frac{V}{V_{0} })

t = -RC ln (1-\frac{V}{V_{0} })        (1)

Next, we can determine the capacitance using the plates' area.

C = ε₀A/d

  = \frac{(8.85*10^{-12))} (20*10^{-4}) }{1*10^{-3} }

  = 18×10⁻¹²F

We can now find the time it takes for the voltage to drop from 100 to 55 V by substituting C, V₀, V, and R values into equation (1)

t = -RC ln (1-\frac{V}{V_{0} })

 = -(1025Ω)(18×10⁻¹² F) ln( 1 - 55/100)

 = 15×10⁻⁹s

= 15 ns

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