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Minchanka [31]
3 years ago
12

Caroline, a piano tuner, suspects that a piano's G4 key is out of tune. Normally, she would play the key along with her G4 tunin

g fork and tune the piano to match, but her G4 tuning fork is missing! Instead, she plays the errant key along with her F4 tuning fork (which has a frequency of 349.2 Hz), displays the resulting waveform on a handheld oscilloscope, and measures a beat frequency of 76.7 Hz. Then, she plays the errant key along with her A4 tuning fork (which has a frequency of 440.0 Hz) and measures a beat frequency of 14.1 Hz.
What frequency is being played by the out-of-tune key ?
a. 363.3 Hz
b. 451.1 Hz
c. 33.9 Hz
d. 272.5 Hz
e. 425.9 Hz
Physics
1 answer:
alexira [117]3 years ago
6 0

Answer:

e. 425.9 Hz

Explanation:

The computation of the frequency is being played by the out-of-tune key is shown below;

Given that

Δf1  = x - 349.2 = 76.7.........(1)

Δf2 = 440 - x  = 14.1......(2)

Now solve (1) and (2)

(440 - x) - x + 349.2 = 14.1 - 76.7

789.2 + (-2x) = -62.6

x = 425.9 Hz

Hence, the frequency is being played by the out-of-tune key is 425.9 Hz

Therefore the option e is correct

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(I) In a ballistic pendulum experiment, projectile 1 results in a maximum height h of the pendulum equal to 2.6 cm. A second pro
Kipish [7]

Answer:

The second projectile was 1.41 times faster than the first.

Explanation:

In the ballistic pendulum experiment, the speed (v) of the projectile is given by:  

v = \frac{m + M}{m} \cdot \sqrt{2gh}

<em>where m: is the mass of the projectile, M: is the mass of the pendulum, g: is the gravitational constant and h: is the maximum height of the pendulum.   </em>

To know how many times faster was the second projectile than the first, we need to take the ratio for the velocities for the projectiles 2 and 1:    

\frac{v_{2}}{v_{1}} = \frac{\frac{m_{2} + M}{m_{2}} \cdot \sqrt{2gh_{2}}}{\frac{m_{1} + M}{m_{1}} \cdot \sqrt{2gh_{1}}}           (1)

<em>where m₁ and m₂ are the masses of the projectiles 1 and 2, respectively, and h₁ and h₂ are the maximum height reached by the pendulum by the projectiles 1 and 2, respectively.  </em>

Since the projectile 1 has the same mass that the projectile 2, we can simplify equation (1):

\frac{v_{2}}{v_{1}} = \frac{\sqrt{h_{2}}}{\sqrt{h_{1}}}  

\frac{v_{2}}{v_{1}} = \frac{\sqrt{5.2 cm}}{\sqrt{2.6 cm}}

\frac{v_{2}}{v_{1}} = 1.41  

Therefore, the second projectile was 1.41 times faster than the first.  

I hope it helps you!

8 0
3 years ago
A firefighting crew uses a water cannon that shoots water at 25.0 m/s at a fixed angle of 53.0° above the horizontal. The fire-f
zysi [14]

Answer:

8.8 m and 52.5 m

Explanation:

The vertical component and horizontal component of water velocity leaving the hose are

v_v = vsin(\alpha) = 25sin(53^0) = 25*0.8 = 19.97 m/s

v_h = vcos(\alpha) = 25cos(53^0) = 25*0.6 = 15 m/s

Neglect air resistance, vertically speaking, gravitational acceleration g = -9.8m/s2 is the only thing that affects water motion. We can find the time t that it takes to reach the blaze 10m above ground level

s = v_vt + gt^2/2

10 = 19.97t - 9.8t^2/2

4.9t^2 - 19.97t + 10 = 0

t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

t= \frac{19.9658877511823\pm \sqrt{(-19.9658877511823)^2 - 4*(4.9)*(10)}}{2*(4.9)}

t= \frac{19.9658877511823\pm14.24}{9.8}

t = 3.49 or t = 0.58

We have 2 solutions for t, one is 0.58 when it first reach the blaze during the 1st shoot up, the other is 3.49s when it falls down

t is also the times it takes to travel across horizontally. We can use this to compute the horizontal distance between the fire-fighters and the building

s_1 = v_ht_1 = 15*0.58 = 8.8 m

s_2 = v_ht_2 = 15*3.49 = 52.5m

8 0
2 years ago
What is the speed of a wave with a wavelength of 3 m and a frequency of .1Hz?
Yuri [45]
We know,
Speed = Frequency * Wavelength 
Speed = 3 * 0.1 m/s   [hertz = 1/sec.]
So, your final answer is 0.3 m/s

Hope this helps!!
6 0
3 years ago
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Leto [7]
The statement: "secondary evidence is the basis for drawing scientific conclusions" is definitely false. Secondary evidence is the body of information obtained to prove the existence of unknown or missing primary evidence. In drawing scientific conclusions, primary evidence is needed.
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2 years ago
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Which best describes the current atomic theory?
ivolga24 [154]
Choice-C is a correct statement.
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