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ruslelena [56]
3 years ago
14

Question 4. A tuning fork ‘A’ produces 6 beats/sec with another fork ‘B’ of un-known frequency. On

Physics
1 answer:
8090 [49]3 years ago
4 0

Clever problem.

We know that the beat frequency is the DIFFERENCE between the frequencies of the two tuning forks.  So if Fork-A is 256 Hz and the beat is      6 Hz, then Fork-B has to be EITHER 250 Hz OR 262 Hz.  But which one is it ?

Well, loading Fork-B with wax increases its mass and makes it vibrate SLOWER, and when that happens, the beat drops to 5 Hz.  That means that when Fork-B slowed down, its frequency got CLOSER to the frequency of Fork-A ... their DIFFERENCE dropped from 6 Hz to 5 Hz.

If slowing down Fork-B pushed it CLOSER to the frequency of Fork-A, then its natural frequency must be ABOVE Fork-A.

The natural frequency of Fork-B, after it gets cleaned up and returns to its normal condition, is 262 Hz.  While it was loaded with wax, it was 261 Hz.

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Answer:

7500 Newtons

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Acceleration of the sportscar= 5m/s^2

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2 years ago
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Rom4ik [11]

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A tray of electronic components contains 15 components, 4 of which are defective. If 4 components are selected, what is the poss
dlinn [17]

Answer:

a) 0.0007326

b) 0.03223

c) 0.2418

d) 0.2418

Explanation:

To find different probabilities for the selection of components among eleven good and four defective components, we will use the Combination.

a) C(4,4) = 1; C(15,4) = 1365

P = \frac{C(4,4)}{C(15,4)} = \frac{1}{1365} = 0.0007326

b) C(4,3) = 4; C(11,1) = 11

P = \frac{C(4,3)*C(11,1)}{C(15,4)} = \frac{4*11}{1365} = 0.03223

c) C(4,2) = 6; C(11,2) = 55

P = \frac{C(4,2)*C(11,2)}{C(15,4)} = \frac{6*55}{1365} = 0.2418

d) C(11,4) = 330

P = \frac{C(11,4)}{C(15,4)} = \frac{330}{1365} = 0.2418

8 0
2 years ago
A stone is dropped from the top of a tower. What is its velocity after 3.0 seconds?
kap26 [50]

For purposes of completing our calculations, we're going to assume that
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The acceleration of gravity on Earth is about 9.8 m/s², directed toward the
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3 seconds after being dropped, a stone is falling at (3 x 9.8) = 29.4 m/s. 

That's the vertical component of its velocity.  The horizontal component is
the same as it was at the instant of the drop, provided there is no horizontal
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5 0
3 years ago
Be-36 what hull type is best for use on ponds, small lakes and calm rivers?
LenaWriter [7]
The hull type that is best for use on ponds, small lakes and calm rivers is Flat Bottom Hull. 
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8 0
3 years ago
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