describe an experiment to show how the frequency of a note emitted by a vibrating string depends on the tension of the string

1 answer:

3 0

Easy !

Take any musical instrument with strings ... a violin, a guitar, etc.

The length of the vibrating part of the strings doesn't change ...
it's the distance from the 'bridge' to the 'nut'.

Pluck any string. Then, slightly twist the tuning peg for that string,
and pluck the string again.

Twisting the peg only changed the string's tension; the length
couldn't change.

-- If you twisted the peg in the direction that made the string slightly
tighter, then your second pluck had a higher pitch than your first one.

-- If you twisted the peg in the direction that made the string slightly
looser, then your second pluck had a lower pitch than the first one.

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A mosquito can fly with a speed of 1.10 m/s with respect to the air. Suppose a mosquito flies east at this speed across a swamp.

sineoko [7]

Answer:

The velocity of Mosquito with respect to earth will be 0.302m/s

Explanation:

V(ma) = 1.10 m/s, east Velocity of mosquito with respect to air

V(ae) = 1.4 m/s at 35° Velocity of air with respect to Earth in west of south direction.

Velocity of Mosquito with respect to earth will be

V(me) = V(ma) + V(ae)

We need to find the mosquito’s speed with respect to Earth in the x direction.

V(x, me) = V(x, ma) + V(x, ae) = V(ma) + V(ae)(cos theta(ae) )

Angle (ae) = –90.0° − 35°=−125°

V(x, me) = 1.10 + (1.4)Cos(-125)

= 1.10 + 1.4(-0.57)

= 1.10 -0.798

= 0.302

So the velocity of Mosquito with respect to earth will be 0.302m/s

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3 years ago

What kind of energy does a spring store a. radiant b. potential c. kinetic d. chemical

vivado [14]

It would be B.Potential energy

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3 years ago

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A boy on a bicycle approaches a brick wall as he sounds his horn at a frequency 400 hz. the sound he hears reflected back from t

Mashutka [201]

As the question is about changing in frequency of a wave for an observer who is moving relative to the wave source, the concept that should come to our minds is "Doppler's effect."

Now the general formula of the Doppler's effect is:
$f = (\frac{g + v_{r}}{g + v_{s}})f_{o}$ -- (A)

Note: We do not need to worry about the signs, as everything is moving towards each other. If something/somebody were moving away, we would have the negative sign. However, in this problem it is not the issue.

Where,
g = Speed of sound = 340m/s.
$v_{r}$ = Velocity of the receiver/observer relative to the medium = ?.
$v_{s}$ = Velocity of the source with respect to medium = 0 m/s.
$f_{o}$ = Frequency emitted from source = 400 Hz.
$f$ = Observed frequency = 408Hz.

Plug-in the above values in the equation (A), you would get:

$408 = ( \frac{340 + v_{r}}{340 + 0})*400$

$\frac{408}{400} = \frac{340 + v_{r}}{340}$

Solving above would give you,
$v_{r}$ = 6.8 m/s

The correct answer = 6.8m/s

7 0

3 years ago

Upon reaching a velocity of 100fps, the pilot of the airplane decides to abort the take off and applies brakes and stops the air

maw [93]

Answer:

$5ft/s^2$

Explanation:

We will use the formula for accelerated motion $v_f^2=v_i^2+2ad$ , which for acceleration it means:

$a=\frac{v_f^2-v_i^2}{2d}$

<em>In our case</em> our initial velocity is $v_i=100ft/s$ and our final velocity is $v_f=0ft/s$ since it comes to rest, this process while traveling a distance $d=1000ft$ , so we have: