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3 years ago

Shoo the fly flaps its wings back and forth 140 times each second. The frequency of the wing flapping is ____ Hz.

Physics

1 answer:

ExtremeBDS [4]3 years ago

5 0

Answer:

140 Hz

Explanation:

Frequency is the reciprocal of period hence f=1/T

Here, the frequency is 140 times per second hence the period will be 1/140= 0.0071428571428 s

Rounded off to 3 significant figures then the period is 0.00714 s or 0.714 ms

Otherwise, the frequency is already given as 140 cycles per second which is equivalent to 140 Hz

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A golfer tees off and hits a golf ball at a speed of 31 m/s and at an angle of 35 degrees. What is the horizontal velocity compo

marishachu [46]

To calculate the horizontal velocity component, use the cosine identity. where cos A = a / h
where A is the angle the golfer hits the ball
a is the horizontal component velocity
h is the speed the golfer hits the ball

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6 0

3 years ago

A 200-Ω resistor is connected in series with a 10-µF capacitor and a 60-Hz, 120-V (rms) line voltage. If electrical energy costs

Vika [28.1K]

Answer:

Cost to leave this circuit connected for 24 hours is $ 3.12.

Explanation:

We know that,

$\mathrm{X}_{\mathrm{c}}=\frac{1}{2 \pi \mathrm{fc}}$

f = frequency (60 Hz)

c= capacitor (10 µF = $10^-6$ )

$\mathrm{X}_{\mathrm{c}}=\text { Capacitive reactance }$

Substitute the given values

$\mathrm{X}_{\mathrm{c}}=\frac{1}{2 \times 3.14 \times 10 \times 10^{-6} \times 60}$

$\mathrm{X}_{\mathrm{c}}=\frac{1}{3.768 \times 10^{-3}}$

$\mathrm{x}_{\mathrm{c}}=265.39 \Omega$

Given that, R = 200 Ω

$X^{2}=R^{2}+X c^{2}$

$X^{2}=200^{2}+265.39^{2}$

$X^{2}=40000+70431.85$

$X^{2}=110431.825$

$x=\sqrt{110431.825}$

X = 332.31 Ω

$\text { Current }(I)=\frac{V}{R}$

$\text { Current }(I)=\frac{120}{332.31}$

Current (I) = 0.361 amps

“Real power” is only consumed in the resistor,

$\mathrm{I}^{2} \mathrm{R}=0.361^{2} \times 200$

$\mathrm{I}^{2} \mathrm{R}=0.1303 \times 200$

$\mathrm{I}^{2} \mathrm{R}=26.06 \mathrm{Watts} \sim 26 \mathrm{watts}$

In one hour 26 watt hours are used.

Energy used in 54 hours = 26 × 24 = 624 watt hours

E = 0.624 kilowatt hours

Cost = (5)(0.624) = 3.12

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Answer:

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