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3 years ago

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Microphones and loudspeakers are used in an auditorium because the sound waves at the stage compared to the sound waves at the b

ack of the auditorium will have different…

1 answer:

MaRussiya [10]3 years ago

6 0

Acoustics (sounds)
Hope the this helps!

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What inertia is present in streched rybber

MrRissso [65]

<h2>You input potential (stored) energy into the rubber band system when you stretched the rubber band back. Because it is an elastic system, this kind of potential energy is specifically called elastic potential energy.</h2><h2>Hope it helps..</h2>

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3 years ago

Given Vout = 17.33 vpp and R1 = 3 kΩ, find the value of RF required to provide Av = 4.33. (Round your answer to 2 decimal places

Olenka [21]

Answer:

The magnitude of $V_{2}$ is 4 V and phase of input voltage is zero

Explanation:

Given:

Output voltage $V_{out} = 17.33$

Resistance $R_{1} = 3$ kΩ

Voltage gain $A_{v} = 4.33$

For finding feedback resistance we use gain equation

Gain equation for non inverting op-amp is given by,

$A_{v} = 1+\frac{R_{f} }{R_{1} }$

$4.33 = 1+ \frac{R_{f} }{3 k }$

$R_{f}$ ≅ 10 kΩ

For finding input voltage we use,

$A_{v} = \frac{V_{out} }{V_{2} }$

$V_{2} = \frac{17.33}{4.33}$

$V_{2} = 4$ V

The Phase of $V_{2}$ is zero because output voltage phase is 360°

Therefore, the magnitude of $V_{2}$ is 4 V and phase of input voltage is zero

7 0

3 years ago

What are four tools you can use to make quantitative observations

aev [14]

Quantitative is how many or a measurement. You can take a measurement with:
a ruler
thermometer
weight scale
stop watch
Hope this helps. <span />

8 0

3 years ago

Read 2 more answers

Question<br> In order for work to be done, what three things are necessary

Misha Larkins [42]

Perseverance, good mind set, and work ethic

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4 years ago

Read 2 more answers

The block slides on a horizontal frictionless surface. The block has a mass of 1.0kg and is pushed 5.0N at 45°. What is the magn

Firlakuza [10]

Answer:

$3.5\:\mathrm{m/s^2}$

Explanation:

Newton's 2nd law is given as $\Sigma F = ma$ .

To find the acceleration in the horizontal direction, you need the horizontal component of the force being applied.

Using trigonometry to find the horizontal component of the force:

$\cos 45^{\circ}=\frac{x}{5},\\\frac{\sqrt{2}}{{2}}=\frac{x}{5},\\x=\frac{5\sqrt{2}}{2}$

Use this horizontal component of the force to solve for for the acceleration of the object:

$\frac{5\sqrt{2}}{2}=1.0\cdot a,\\a=\frac{5\sqrt{2}}{2}\approx \boxed{3.5\:\mathrm{m/s^2}}$

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2 years ago