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Licemer1 [7]
3 years ago
9

A 1056-hertz tuning fork is struck at the same time as a note on the piano and you hear 2 beats/second. You tighten the piano st

ring very slightly and then hear 3 beats/second. What is the frequency of the piano string?
Physics
1 answer:
elena-14-01-66 [18.8K]3 years ago
6 0

Answer:

The frequency of the piano string is <em>1059 Hz</em>.

Explanation:

The frequency beat (fb), 2 beats/second, is the absolute difference between the frequency of the tuning fork (1056 Hz) and the frequency of the piano string.

As the piano string gets tightened, the frequency beat becomes 3 beats/second.

Therefore,

fb = fb = fpiano - ftuning fork\\ 3 Hz=fpiano-1056Hz\\ fpiano=1056Hz+3Hz\\ fpiano=1059Hz

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Ladybug walks 10 cm forward and 5 cm backwards in 20 seconds what is the average speed of ladybug what is the average velocity
lbvjy [14]

Answer:

Average speed = 0.0075 m/s

Average velocity = 0.0025 m/s along forward direction

Explanation:

Speed is the ratio of distance and time and velocity is the ratio of displacement and time.

Distance traveled = 10 + 5 = 15 cm = 0.15 m

Displacement = 10 - 5 = 5 cm = 0.05 m

Time = 20 seconds

\texttt{Average speed = }\frac{\texttt{Distance}}{\texttt{Time}}\\\\\texttt{Average speed = }\frac{0.15}{20}=7.5\times 10^{-3}m/s\\\\\texttt{Average velocity = }\frac{\texttt{Displacement}}{\texttt{Time}}\\\\\texttt{Average velocity = }\frac{0.05}{20}=2.5\times 10^{-3}m/s

Average speed = 0.0075 m/s

Average velocity = 0.0025 m/s along forward direction

3 0
3 years ago
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During an earthquake, where is the greatest motion felt on the surface?
dalvyx [7]
That would be the epicenter. 
5 0
3 years ago
Anyone knows this please help me
olasank [31]

Answer:

TRUE

Explanation:

3 0
3 years ago
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Please help on this one
Sphinxa [80]

Since this is a projectile motion problem, break down each of the five kinematic quantities into x and y components. To find the range, we need to identify the x component of the displacement of the ball.

Let's break them down into components.

                X                                Y

v₁     32 cos50 m/s           32 sin50 m/s

v₂     32 cos50 m/s                    ?

Δd             ?                                0

Δt              ?                                ?

a                0                         -9.8 m/s²


Let's use the following equation of uniform motion for the Y components to solve for time, which we can then use for the X components to find the range.

                           Δdy = v₁yΔt + 0.5ay(Δt)²

                                0 = v₁yΔt + 0.5ay(Δt)²

                                0 = Δt(v₁ + 0.5ayΔt), Δt ≠ 0

                                0 = v₁ + 0.5ayΔt

                                0 = 32sin50m/s + 0.5(-9.8m/s²)Δt

                                0 = 2<u>4</u>.513 m/s - 4.9m/s²Δt

                 -2<u>4</u>.513m/s = -4.9m/s²Δt

-2<u>4</u>.513m/s  ÷ 4.9m/s² = Δt

                          <u>5</u>.00s = Δt


Now lets put our known values into the same kinematic equation, but this time for the x components to solve for range.

Δdₓ = v₁ₓΔt + 0.5(a)(Δt)²

Δdₓ = 32cos50m/s(<u>5</u>.00s) + 0.5(0)(<u>5</u>.00)²

Δdₓ = 32cos50m/s(<u>5</u>.00s)

Δdₓ = 10<u>2</u>.846


Therefore, the answer is A, 102.9m. According to significant digit rules, neither would be correct, but 103m is the closest to 102.9m so I guess that is what it is.






6 0
3 years ago
The speed limit on many roads in a town 13.5 m/s outside schools this is limit is often reduced by one third
pochemuha

Answer:

9 m/s

Explanation:

13.5 × \frac{2}{3} = 9

5 0
2 years ago
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