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Yuri [45]
3 years ago
5

A piano tuner is tuning a piano when he discovers that the G above middle C is vibrating with a higher frequency than his G tuni

ng fork, which vibrates at 392.0 Hz. He plays the piano key and tuning fork at the same time and hears a beat frequency of 2.0 Hz.
a. What is the frequency of the G on Ms. Carlton's piano?
b. If it had been a lower frequency, creating a beat frequency of 4.0 Hz, what would it have been?
Physics
1 answer:
kotegsom [21]3 years ago
6 0

Answer:

Tuning a piano is the process of adjusting the tension of its strings, thereby altering their pitch, or frequency of vibration, by slightly turning the tuning pins to which they're attached, so that each string sounds pleasingly in harmony with every other string.

<h2>Hopefully u will satisfy with my answer of ur question..!!</h2>

<h2>Have a nice day ahead dear..!!</h2>
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The question is on the picture.
Jet001 [13]
Between noon and 2 pm, the amount of water in the rain gauge decreased.
This can be caused by evaporation, which turns water into water vapor.
Precipitation would increase the amount of rain water in the gauges, not decrease it.
Condensation occurs after evaporation but wouldn't decrease the water in the gauges by itself.
Runoff is when water on land drains into water sources such as lakes, rivers, oceans, etc. 
So the answer is A. evaporation.
5 0
3 years ago
Read 2 more answers
Decide which of the following day-to-day activities might be considered aerobic exercise?
erastova [34]

Answer:

3. Mowing the lawn

Explanation:

Aerobic exercise is any type of cardiovascular conditioning.

Mowing the lawn is the only thing that is actively moving.

3 0
3 years ago
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A 10-kg cart moving at 5 m/s collides with a 5-kg cart at rest and causes it to move 10 m/s. Which principle explains the result
Virty [35]

Hello!

A 10-kg cart moving at 5 m/s collides with a 5-kg cart at rest and causes it to move 10 m/s. Which principle explains the result? A) law of differential mass B) law of conservation of momentum C) law of unequal forces D) law of accelerated collision

We have the following data¹:

ΔP (momentum before impact) = ?  

mA (mass) = 10 kg

vA (velocity) = 5 m/s

mB (mass) = 5 kg

vB (velocity) = 0 m/s

Solving:

ΔP = mA*vA + mB*vB

ΔP = 10 kg*5 m/s + 5 kg*0 m/s

ΔP = 50 kg*m/s + 0 kg*m/s

Δp = 50 kg*m/s ← (momentum before impact)

We have the following data²:

ΔP (momentum after impact) = ?  

mA (mass) = 10 kg

vA (velocity) = 0 m/s

mB (mass) = 5 kg

vB (velocity) = 10 m/s

Solving:

Δp = mA*vA + mB*vB

Δp = 10 kg*0 m/s + 5 kg*10 m/s

Δp = 0 kg*m/s + 50 kg*m/s

Δp = 50 kg*m/s ← (momentum after impact)

*** Then, which principle explains the result ?

Law of conservation of momentum, <u>since the total momentum of body A and B before impact is equal to the total momentum of body A and B after impact.</u>

Note:  Bodies of different masses and velocities may have the same kinetic energy, if proportionality between the units is maintained it can occur that they have the same kinetic energy.

Answer:

B) law of conservation of momentum

_______________________

\bf\red{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}\:\:\ddot{\smile}

5 0
3 years ago
A weather balloon is inflated to a volume of 26.8 LL at a pressure of 744 mmHgmmHg and a temperature of 31.2 ∘C∘C. The balloon r
elena-s [515]

Answer:

43.96 L

Explanation:

We are given that

V_1=26.8 L

P_1=744mm Hg

T_1=31.2^{\circ} C=31.2+273=304.2K

P_2=385mmHg

T_2=-14.8^{\circ}=273-14.8=258.2K

We know that

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

V_2=\frac{P_1V_1T_2}{P_2T_1}

Substitute the values

V_2=\frac{744\times 26.8\times 258.2}{385\times 304.2}

V_2=43.96 L

Hence, the volume of balloon at -14.8 degree Celsius=43.96 L

5 0
3 years ago
A pulsar is a rapidly rotating neutron star that emits a radio beam the way a lighthouse emits a light beam. We receive a radio
DanielleElmas [232]

Answer:

\alpha =-2.2669642\times^{-10}rad/s^2

Explanation:

Angular acceleration is defined by \alpha =\frac{\Delta \omega}{\Delta t}=\frac{\omega_f-\omega_i}{\Delta t}

Angular velocity is related to the period by \omega=\frac{2\pi}{T}

Putting all together:

\alpha =\frac{\frac{2\pi}{T_f}-\frac{2\pi}{T_i}}{\Delta t}=\frac{2\pi}{\Delta t}(\frac{1}{T_f}-\frac{1}{T_i})

Taking our initial (i) point now and our final (f) point one year later, we would have:

\Delta t=1\ year=(365)(24)(60)(60)s=31536000&#10;s

T_i=0.0786s

T_f=0.0786s+7.03\times10^{-6}s

So for our values we have:

\alpha =\frac{2\pi}{\Delta t}(\frac{1}{T_f}-\frac{1}{T_i})=\frac{2\pi}{31536000s}(\frac{1}{0.0786s+7.03\times10^{-6}s}-\frac{1}{0.0786s})=-2.2669642\times^{-10}rad/s^2

Where the minus sign indicates it is decelerating.

8 0
3 years ago
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