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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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A bird watcher meanders through the woods, walking 1.93 km due east, 1.03 km due south, and 3.84 km in a direction 52.8 ° north
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Answer:

Magnitude of displacement = 2.07 km

Magnitude of average velocity = 1.17 kmph

Explanation:

Let east represent positive x axis and north represent positive y axis.

A bird watcher meanders through the woods, walking 1.93 km due east, 1.03 km due south, and 3.84 km in a direction 52.8 ° north of west.

1.93 km due wast

           s ₁ = 1.93 i km

1.03 km due south

           s₂ = -1.03 j km

3.84 km in a direction 52.8 ° north of west

           s₃ = -3.84 cos 52.8 i + 3.84 sin 52.8 j = -2.32 i + 3.06 j km

Total displacement

          s = s ₁+  s₂+ s₃ = 1.93 i - 1.03 j -2.32 i + 3.06 j = -0.39 i + 2.03 j

  Magnitude of displacement, =\sqrt{(-0.39)^2+2.03^2}=2.07km

Time taken = 1.771 hour

Magnitude of average velocity, =\frac{2.07}{1.771}=1.17km/hr

7 0
3 years ago
Arrange these source charges in order from least to greatest magnitude.
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C ,A ,B ,D

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Five hundred joules of heat are added to a closed system. The initial internal energy of the system is 87 J, and the final inter
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We can solve the problem by using the first law of thermodynamics:

\Delta U= Q-W

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\Delta U is the variation of internal energy of the system

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W is the work done by the system

In this problem, the variation of internal energy of the system is

\Delta U=U_f-U_i=134 J-87 J=47 J

While the heat added to the system is

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therefore, the work done by the system is

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A man wishes to pull a crate 15m across a rough floor by exerting a force of 100 N. The
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Answer:

option (E) is correct.

Explanation:

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force, f = 100 N

Coefficient of friction, = 0.25

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So, net force F = f - friction force

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For minimum work, the angle should be maximum.

So, the value of θ is 76°.

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