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kumpel [21]
3 years ago
14

Suppose a piano tuner hears 2 beats per second when listening to the combined sound from her tuning fork and the piano note bein

g tuned. After slightly tightening the string, she hears 1 beat per second. Should she loosen or should she further tighten the string?
Physics
1 answer:
maksim [4K]3 years ago
6 0

Answer:

further tightening is required.

Explanation:

The beat created / sec = difference of frequencies

Initial beat heard = 2

so difference of frequencies = 2

after tightening beat heard  = 1

difference  of frequencies decreases because frequency of tuning fork was higher than piano sound.

on further tightening  difference decreases because tightening increases the frequency  of piano hence   further  tightening is required for resonance.

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4 0
2 years ago
A boy is whirling a stone around his head by means of a string. The string makes one complete revolution every second; and the m
Damm [24]

Answer

given,                                                

Tension of string is F                                                  

velocity is increased and the radius is not changed.      

the string makes two complete revolutions every second

consider the centrifugal force acting on the stone          

  = \dfrac{mv^2}{r}                          

now centrifugal force is balanced by tension

T =\dfrac{mv^2}{r}                                

From the above expression we can clearly see that tension is directly proportional to velocity and inversely proportional to radius.

When radius is not changing velocity is increasing means tension will also increase in the string.

8 0
3 years ago
A student pulls horizontally on a 12 kg box, which then moves horizontally with an acceleration of 0.2 m/s^2. If the student use
polet [3.4K]
The net force of the object is equal to the force applied minus the force of friction. 
                         Fnet = ma = F - Ff
                           12 kg x 0.2 m/s² = 15 N - Ff
The value of Ff is 12.6 N. This force is equal to the product of the normal force which is equal to the weight in horizontal surface and the coefficient of friction.
                             Ff = 12.6 N = k(12 kg)(9.81 m/s²)
The value of k is equal to 0.107. 
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3 years ago
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Answer:

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3 years ago
A contestant in a winter games event pulls a 36.0 kg block of ice across a frozen lake with a rope over his shoulder as shown in
Novay_Z [31]

(a) The minimum force F he must exert to get the block moving is 38.9 N.

(b) The acceleration of the block is 0.79 m/s².

<h3>Minimum force to be applied </h3>

The minimum force F he must exert to get the block moving is calculated as follows;

Fcosθ = μ(s)Fₙ

Fcosθ = μ(s)mg

where;

  • μ(s) is coefficient of static friction
  • m is mass of the block
  • g is acceleration due to gravity

F = [0.1(36)(9.8)] / [(cos(25)]

F = 38.9 N

<h3>Acceleration of the block</h3>

F(net) = 38.9 - (0.03 x 36 x 9.8) = 28.32

a = F(net)/m

a = 28.32/36

a = 0.79 m/s²

Thus, the minimum force F he must exert to get the block moving is 38.9 N.

The acceleration of the block is 0.79 m/s².

Learn more about minimum force here: brainly.com/question/14353320

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2 years ago
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