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Inga [223]
3 years ago
7

Two of the three tuning forks have known frequencies. When a 510 Hz. fork and the unknown fork are struck together, four beats p

er second are heard. When a 505 Hz. fork is struck with the unknown fork, beats are produced that are too numerous to count accurately. What is the frequency of the unknown fork?
Physics
1 answer:
Vikki [24]3 years ago
3 0
4 Hz is the difference between 510 and the unknown. 
<span>Therefore the unknown is either 514 or 506. </span>

<span>If 505 and 506 were struck together, the diff would be 1 Hz </span>
<span>If 505 and 514 were struck toghether, the diff would be 9 Hz which would be difficult to count accurately, </span>

<span>Therefore the unknown is 514</span>
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Answer:

2.When they reach the bottom of the fall

Explanation:

The potential energy of the waterfall is maximum at the maximum height and decreases with decrease in height. Based on the law of conservation of mechanical energy, as the potential energy of the water fall is decreasing with  decrease in height of the fall, its kinetic energy will be increasing and the kinetic energy will be maximum at zero height (bottom of the fall).

Thus, the correct option is "2" When they reach the bottom of the fall

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2 years ago
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scZoUnD [109]

Answer:

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Explanation:

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2 years ago
Calculate the electric field at the center of a square
pantera1 [17]

Answer:

E_y=1175510.2\ N.C^{-1}

The Magnitude of electric field is in the upward direction as shown directly towards the charge q_1.

Explanation:

Given:

  • side of a square, a=52.5\ cm
  • charge on one corner of the square, q_1=+45\times 10^{-6}\ C
  • charge on the remaining 3 corners of the square,q_2=q_3=q_4=-27\times 10^{-6}\ C

<u>Distance of the center from each corners</u>=\frac{1}{2} \times diagonals

diagonal=\sqrt{52.5^2+52.5^2}

diagonal=74.25\cm=0.7425\ m

∴Distance of center from corners, b=0.3712\ m

Now, electric field due to charges is given as:

E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}

<u>For charge q_1 we have the field lines emerging out of the charge since it is positively charged:</u>

E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}

  • E_1=2938775.5\ N.C^{-1}

<u>Force by each of the charges at the remaining corners:</u>

E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}

  • E_2=E_3=E_4=1763265.3\ N.C^{-1}

<u> Now, net electric field in the vertical direction:</u>

E_y=E_1-E_4

E_y=1175510.2\ N.C^{-1}

<u>Now, net electric field in the horizontal direction:</u>

E_y=E_2-E_3

E_y=0\ N.C^{-1}

So the Magnitude of electric field is in the upward direction as shown directly towards the charge q_1.

8 0
3 years ago
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Answer: A

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f=50*300=15000 Hz = 15kHz.

Have a great day! <3

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1 year ago
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Whitepunk [10]

Answer:

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The use of bearing surfaces that are themselves sacrificial, such as low shear materials, of which lead/copper journal bearings are an example

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