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klio [65]
3 years ago
6

Harmonics problem. A square wave of frequency f contains harmonics (sine waves) at f, 3f, 5f, 7f, ... . Suppose a system respond

s to frequencies in the range 20-40 kHz but is insensitive outside of this range.EXAMPLE:Imagine applying a square wave with f = 30 kHz to this system. What is the shape of the response?[Answer: a 30 kHz sine wave. ]Imagine applying a square wave with f = 10 kHz to this system.What is the shape of the response?A. A 10 kHz sine waveB. A combination of 20, 30 and 40 kHz sine wavesC. A 30 kHz sine waveD. No response
Physics
1 answer:
ira [324]3 years ago
4 0

Answer:

B

Explanation:

A square of frequency of consists of the infinite sum of sine waves, whose frequencies are the odd multiples of the main frequency f i.e f, 3f,5f, 7f, ... . Given that the range of frequencies, to which the system responds is 20-40 kHz, for a square wave of frequency 10kHz we need to look for the harmonics whose frequencies are in the systems' respond range, which are the  harmonics of 20, 30 and 40 kHz

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A playground ride consists of a disk of mass M = 49 kg and radius R = 1.7 m mounted on a low-friction axle. A child of mass m =
HACTEHA [7]

Answer:

The angular speed is 0.83 rad/s.

Explanation:

Given that,

Mass of disk M=49 kg

Radius = 1.7 m

Mass of child m= 29 kg

Speed = 2.6 m/s

Suppose if the disk was initially at rest , now how fast is it rotating

We need to calculate the angular speed

Using conservation of momentum

m\omega_{i}=(mr^2+\dfrac{Mr^2}{2})\omega_{f}

mvR=(mr^2+\dfrac{Mr^2}{2})\omega

Put the value into the formula

29\times2.6\times1.7=(29\times1.7^2+\dfrac{49\times1.7^2}{2})\omega_{f}

\omega_{f}=\dfrac{29\times2.6\times1.7}{(29\times1.7^2+\dfrac{49\times1.7^2}{2})}

\omega_{f}=0.83\ rad/s

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4 0
3 years ago
A stone is tossed horizontally from the highest point of a 95 m building and lands 105 m from the base of the building. Ignore a
aalyn [17]

Answer:

A) t = 4.40 s , B)   v = 23.86 m / s ,  c)  v_y = - 43.12 m / s , D)  v = 49.28 m/s

Explanation:

This is a projectile throwing exercise,

A) To know the time of the stone in the air, let's find the time it takes to reach the floor

          y = y₀ + v_{oy} t - ½ g t²

as the stone is thrown horizontally  v_{oy} = 0

          y = y₀ - ½ g t²

          0 = y₀ - ½ g t²

          t = √ (2 y₀ / g)

          t = √ (2 95 / 9.8)

          t = 4.40 s

B) what is the horizontal velocity of the body

          v = x / t

          v = 105 / 4.40

          v = 23.86 m / s

C) The vertical speed when it touches the ground

          v_y = v_{oy} - g t

          v_y = 0 - 9.8 4.40

          v_y = - 43.12 m / s

the negative sign indicates that the speed is down

D) total velocity just hitting the ground

          v = vₓ i ^ + v_y j ^

          v = 23.86 i ^ - 43.12 j ^

Let's use Pythagoras' theorem to find the modulus

          v = √ (vₓ² + v_y²)

          v = √ (23.86² + 43.12²)

           v = 49.28 m / s

we use trigonometry for the angle

          tan θ = v_y / vₓ

          θ = tan⁻¹ (-43.12 / 23.86)

            θ = -61

7 0
4 years ago
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daser333 [38]
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so x = - 28 m</span>
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4 years ago
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