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vladimir1956 [14]
3 years ago
8

3. Do you have an issue with the NSA monitoring citizens’ phone calls, text messages, emails, etc.?

Physics
1 answer:
goblinko [34]3 years ago
8 0
<span> If You Have Nothing to Hide, You Have Nothing to Fear</span>
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Two loudspeakers are placed side by side and driven by the same source at 500 Hz. A listener is positioned in front of the two s
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Answer:

0.68 m

Explanation:

We know that the speed of sound in air is a product of frequency and wavelength. Taking speed of sound in air as 340 m/s

V=frequency*wavelength

Then wavelength is given by 350/500=0.68 m

Therefore, to repeat constructive interference at the listener's ear, a distance of 0.68 m is needed

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3 years ago
A 160.-kilogram space vehicle is traveling along a straight line at a constant speed of 800. Meters per second. The magnitude of
natulia [17]

Answer:

Zero

Explanation:

As force acting on the body is equal to the product of mass and acceleration.

Acceleration is equal to rate of change in velocity.

Here velocity is constant so acceleration is zero.

It means the net force acting on the vehicle is zero.

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3 years ago
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Apply the impulse-momentum relation and the work-energy theorem to calculate the maximum value of t if the cake is not to end up
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What is the current in a 160V circuit if the resistance is 2Ω?<br> V=<br> I=<br> R=
Alex787 [66]

Explanation:

v = IR

v= 160 R = 2

<u>160</u> = <u>2I</u>

2 2

I = 80A

4 0
3 years ago
A record player turntable initially rotating at 3313 rev/min is braked to a stop at a constant rotational acceleration. The turn
Rus_ich [418]

Answer:

(A) It will take 22 sec to come in rest

(b) Work done for coming in rest will be 0.2131 J              

Explanation:

We have given the player turntable initially rotating at speed of 33\frac{1}{3}rpm=33.333rpm=\frac{2\times 3.14\times 33.333}{60}=3.49rad/sec

Now speed is reduced by 75 %

So final speed \frac{3.49\times 75}{100}=2.6175rad/sec

Time t = 5.5 sec

From first equation of motion we know that '

\alpha =\frac{\omega -\omega _0}{t}=\frac{2.6175-3.49}{4}=-0.158rad/sec^2

(a) Now final velocity \omega =0rad/sec

So time t to come in rest  t=\frac{0-3.49}{-0.158}=22sec

(b) The work done in coming rest is given by

\frac{1}{2}I\left ( \omega ^2-\omega _0^2 \right )=\frac{1}{2}\times 0.035\times (0^2-3.49^2)=0.2131J

4 0
3 years ago
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