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Dimas [21]
3 years ago
15

A horse runs a race on a oval track and starts and ends at the same point. He runs one mile track in 7 minutes.

Physics
1 answer:
Sergeeva-Olga [200]3 years ago
4 0

Answer:

speed = 8.57mph while v = 0 mph

Explanation:

The horse is running 1 mile in 7 minutes. Then for a total of 1 hour, 60 minutes, the horse should be able to run a distance of

\frac{1*60}{7} = \frac{60}{7} = 8.57mph

So the horse speed is 8.57 miles per 60 minutes, or 8.57 mph (mile per hour). But if he ends up at the same point as before int he oval track, one can argue that the displacement is 0, and so is its velocity.

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A) Force

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In a series RLC resonance circuit, the resonance frequency f0 = 700 kHz. The resistor R = 10 Ohm. The specified bandwidth (BW) s
sladkih [1.3K]

Answer:

  • quality factor (Q) = 69.99
  • inductor = 1.591 x 10⁻⁴ H
  • capacitor = 3.248 x 10⁻¹⁰ F

Explanation:

Given;

resonance frequency (F₀) = 700 kHz

resistor, R =  10 Ohm

bandwidth (BW) = 10 kHz

bandwidth (BW)  = \frac{R}{2\pi L}

BW = \frac{R}{2\pi L}

make L (inductor) the subject of the formula

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F_o =\frac{1}{2\pi\sqrt{LC} } \\\\\sqrt{LC} = \frac{1}{2\pi F_o} \\\\LC = \frac{1}{4\pi ^2F_o^2}= \frac{1}{4\pi ^2(700,000)^2} = 5.168*10^{-14}

make C (capacitor)  the subject of the formula

C = \frac{5.168*10^{-14}}{1.591*10^{-4}} = 3.248*10^{-10} \ F = \ 3.248*10^{-4} \ \mu F

quality factor (Q) = \frac{1}{R} \sqrt{\frac{L}{C}} \ = \frac{1}{10} \sqrt{\frac{1.591*10^{-4}}{3.248*10^{-10}}}=69.99

quality factor (Q) =  69.99

5 0
4 years ago
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