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o-na [289]
3 years ago
5

Find the frequency of a wave produced by a generator that emits 100 pulses in 2.0 s.

Physics
1 answer:
kobusy [5.1K]3 years ago
7 0

The frequency of the wave is 50 Hz

Explanation:

The frequency of the wave is defined as the number of cycles per second of the wave:

f=\frac{N}{t}

where

N is the number of cycles completed in a time t.

Frequency is measured in Hertz (Hz).

In this problems, the wave has

N = 100 pulses

in

t = 2.0 s

Therefore, its frequency is

f=\frac{100}{2.0}=50 Hz

Learn more about waves and frequency here:

brainly.com/question/5354733

brainly.com/question/9077368

#LearnwithBrainly

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A complete circuit contains two parallel connected devices and a generator for providing the electromotive force. The resistance
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We are asked to solve and determine the magnitude of the current flowing through the first device. In order for us to have a better understanding of the problem, we can refer to the attached picture which contains electric circuit diagram. Since it the problem we are already given with an electromotive source or the voltage supply and since the two resistance is in parallel, it would clearly mean that the voltage drop in each resistance is just the same. The resistance 1 uses the 40 volts at the same time the resistance 2 uses 40 volts also. Solving further for the current, we can apply Ohm's law which V = IR where "V" represents the voltage, the "I" represents the current and "R" represents the resistance.

Such as the solution in obtaining current is shown below:
I = V / R, substitute values we have it
I = 40 volts / 1208 ohms
I = 0.0331 Amperes

Therefore, the current flowing in the first device is 0.033 Amperes or 33 milliAmperes.

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A hill that has a 15.5% grade is one that rises 15.5 m vertically for every 100.0 m of distance in the horizontal direction. At
Andreas93 [3]

The angle of incline of the hill above the horizontal is 8.81°.

Since the hill has a 15.5% grade is one that rises 15.5 m vertically for every 100.0 m of distance in the horizontal direction.

<h3>Tangent of the angle of the incline of the hill,</h3>

The tangent of the angle of the incline of the hill, Ф is

tanФ = vertical rise/horizontal distance = grade of hill

Now, the vertical rise = 15.5 m and the horizontal distance = 100.0 m

So, substituting the values of the variables into the equation, we have

tanФ = vertical rise/horizontal run

tanФ = 15.5 m/100.0 m

tanФ = 0.155

<h3>Angle of incline of the hill</h3>

Taking inverse tan of both sides, we have

Ф = tan⁻¹(0.155)

Ф = 8.81°

So, the angle of incline of the hill above the horizontal is 8.81°.

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A 44-turn rectangular coil with length ℓ = 17.0 cm and width w = 8.10 cm is in a region with its axis initially aligned to a hor
Mumz [18]

Answer:

The maximum induced emf in the rotating coil  = 29.66V

The induced emf in the rotating coil when (t = 1.00 s) = 26.66V

The maximum rate of change of the magnetic flux through the rotating coil = 0.674Wb/s

Explanation:

Lets state the parameters we are being given right from the question:

Number of rectangular coil, (N) = 44

Length of Coil, l =17cm in meters we have; (l) = 17 × 10⁻² m

Width of Coil, w =8.10cm in meters we have; (w) = 8.10 × 10⁻² m

Magnitude of Uniform Magnetic Field (B) = 767mT= 765 × 10⁻³ T

Angular Speed of Coil, (ω) = 64 rad/s

(a)

To calculate the induced emf in the rotating cell,we can use the formula:

emf = NBAωsin(ωt)

For maximum induced emf, the value of sin(ωt) will be 1

emf_max = NBAω ; if (A = l × w) , we have:

emf_max  = NB(l × w)ω

subsitituting the parameters into the above equation; we have:

emf_max  = 44 × 765 × 10⁻³ ( 17 × 10⁻² × 8.10 × 10⁻² ) × 64

= 29.66V

(b)

At t = 1s, the induced emf is calculated as:

emf = NBAωsin(ωt)

substituting the parameters into the equation, we have:

emf =   44 × 765 × 10⁻³ ( 17 × 10⁻² × 8.10 × 10⁻² ) × 64 × sin (64 × 1)

=26.66V

(c)

To calculate the maximum rate of change of the magnetic flux through the rotating coil; we need to reflect on the equation for the maximum induced emf in terms of magnetic flux.

i.e emf_max = N\frac{d∅}{dt}

since emf_max = 29.66 and N = 44; we have:

29.66 =  44\frac{d∅}{dt}

\frac{d∅}{dt} = \frac{29.66}{44}

= 0.674 Wb/s

5 0
3 years ago
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