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3 years ago

Answer right I will cashapp $5

Physics

2 answers:

Sholpan [36]3 years ago

6 0

Answer:

The original analog sound waves are perfect but the copy is not because most analog waves have wavelengths that are too long to reproduce in copies

The original digital waves and the copy are perfect because with digital waves no information gets lost in the transfer of data, and there is no additional noise added to the copy.

Analog refers to audio recorded using methods that replicate the original sound waves. Digital audio is recorded by taking samples of the original sound wave at a specified rate.

Explanation:

i dont need $5

i wrote this using information from go ogle and some from brai nly. I hope it helps.

miss Akunina [59]3 years ago

5 0

Analog transfer information quicker, example electromagnetic waves which are used in our modern day phones.

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What is the relation between force and acceleration

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3 years ago

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Katyanochek1 [597]

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chubhunter [2.5K]

Complete Question

The complete quetion is shown on the first uploaded image

Answer:

Explanation:

From the question we are told that

The mass of the fly is $m_f = 11 mg = 11*10^{-3} g = 1.1*10^{-5} \ kg$

The extension of the web is $e= 4.00 \ mm = 0.004 \ m$

The spring constant is mathematically evaluated as

$k = \frac{mg}{e}$

substituting values

$k = \frac{1.1 *10^{-5} *9.8}{0.004}$

$k = 0.027 \ N/m$

The frequency of vibration is

$f = \frac{1}{2 \pi} \sqrt{\frac{k}{m} }$

substituting values

$f = \frac{1}{2 * 3.142 } \sqrt{\frac{0.027}{1.1*10^{-5}} }$

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8 0

3 years ago

7) A crazy cat (yes, this is redundant) is running along the roof of a 60 m tall building. The cat is moving at a constant veloc

gregori [183]

Answer:

The distance from the base of the building to the landing site is 154 m.

The total flight time is 3.5 s.

At the moment of impact, the velocity vector of the cat is v = (44 m/s, -34.3 m/s) and its magnitude is 55.8 m/s.

Explanation:

The equations for the position and velocity vectors of the cat are as follows:

r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)

v = (v0x, v0y + g · t)

Where:

r = position vector of the cat at time t.

x0 = initial horizontal position.

v0x = initial horizontal velocity.

t = time.

y0 = initial vertical position.

v0y = initial vertical velocity

g = acceleration due to gravity (-9.8 m/s² considering the upward position as positive).

v = velocity vector of the cat at time t.

Please, see the attached figure for a better understanding of the problem. Notice that the origin of the frame of reference is located at the launching point so that x0 and y0 = 0. In a horizontal launch, initially there is no vertical velocity, then, v0y = 0.

When the cat reaches the ground, the position vector of the cat will be r1 (see figure). The vertical component of r1 is -60 m and the horizontal component will be the horizontal distance traveled by the cat (r1x). Then, using the equation of the y-component of the position vector, we can obtain the time of flight and with that time we can obtain the horizontal distance traveled by the cat:

r1y = y0 + v0y · t + 1/2 · g · t²

-60 m = 0 m + 0 m/s · t - 1/2 · 9.8 m/s² · t²

- 60 m = -4.9 m/s² · t²

-60 m / - 4.9 m/s² = t²

t = 3.5 s

The cat reaches the ground in 3.5 s

Now, we can calculate the horizontal component of r1:

r1x = x0 + v0 · t

r1x = 0 m + 44 m/s · 3.5 s

r1x = 154 m

The distance from the base of the building to the landing site is 154 m.

The total flight time was already calculated and is 3.5 s.

The velocity vector of the cat when it reaches the ground will be:

v = (v0x, v0y + g · t)

v = (44 m/s, 0 m/s - 9.8 m/s² · 3.5 s)

v = (44 m/s, -34.3 m/s)

The magintude of the vector "v" is calculated as follows:

$|v| = \sqrt{(44 m/s)^{2}+(-34.3 m/s)^{2}} = 55.8 m/s$

At the moment of impact, the velocity vector of the cat is v = (44 m/s, -34.3 m/s) and its magnitude is 55.8 m/s.

6 0

3 years ago