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S_A_V [24]
3 years ago
14

A 240 cm length of string has a mass of 2.5 g. It is stretched with a tension of 8.0 N between fixed supports. (a) What is the w

ave speed for this string? (b) What is the lowest resonant frequency of this string?
Physics
1 answer:
kotykmax [81]3 years ago
3 0

Answer:

a.87.6 m/s

b.18.25 Hz

Explanation:

the equation for the wave speed is expressed as

v=\sqrt{\frac{Tl}{m}}\\

where v is the speed,

           T is the tension in Newton

           l is the length

and      m is the mass

Now since

T=8.0N\\m=2.5g=0.0025kg\\l= 240cm=2.40cm\\

by substituting values into the equation, we have

v=\sqrt{\frac{8*2.40}{0.0025} } \\v=87.6m/s\\

b. the expression for the frequency is giving as

Frequency,f=\frac{v}{2l} \\f=\frac{87.6}{2*2.40} \\f=18.25Hz\\.

Note we 2L as the wavelength because we solving for the fundamental frequency as stated in the question.

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During a tennis volley, a ball that arrives at a player at 40 m/s is struck by the racquet and returned at 40 m/s. The other pla
Butoxors [25]

Answer:Racquet force is twice of Player force

Explanation:

Given

ball arrives at a speed of u=-40\ m/s

ball returned with speed of v=40\ m/s

average Force imparted by racquet on the ball is given by

F_{racquet}=\frac{m(v-u)}{\Delta t}

where m=mass\ of\ ball

\Delta t=time of contact of ball with racquet

F_{racquet}=\frac{m(40-(-40))}{\Delta t}

F_{racquet}=\frac{80m}{\Delta t}-----1

When it land on the player hand its final velocity becomes zero and time of contact is same as of racquet

F_{player}=\frac{m(0-40)}{\Delta t}

F_{player}=\frac{-40m}{\Delta t}-----2

From 1 and 2 we get

F_{racquet}=-2F_{player}

Hence the magnitude of Force by racquet is twice the Force by player

5 0
4 years ago
Describe two ways unbalanced forces help you in your day to<br> day life.
Alexxandr [17]

Answer:

  1. we need unbalance force to lift objects
  2. we need unbalance force to drag objects
6 0
3 years ago
HELP BRANLIEST
Vsevolod [243]

Explanation:

- Newton's first law of motion:

"An object at rest (or in uniform motion) remains at rest (or in uniform motion) unless acted upon an unbalanced force

In this situation, we can apply Newton's first law to the keys of the keyboard that are not hit by the fingers of the man. In fact, as no force act on the keys, they remain at rest.

- Newton's second law of motion:

"The acceleration experienced by an object is proportional to the net force exerted on the object; mathematically:

F=ma

where F is the net force, m is the mass of the object, and a its acceleration"

In this case, we can apply Newton's second law to the keys of the keyboard that are hit by the man: in fact, as they are hit, they experience a downward force, and therefore they experience a downward acceleration.

"Newton's third law of motion:

"When an object A exerts a force on an object B (action force), then object B exerts an equal and opposite force on object A (reaction force)"

Here We can apply Newton's third law to the pair of objects finger-key: in fact, as the finger apply a force on the key (action force), then the key exerts a force back on the finger (reaction force), equal and opposite.

3 0
3 years ago
Select the correct answer.
r-ruslan [8.4K]

Answer:

That would be B. Hope this helps!

Explanation:

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3 0
3 years ago
Read 2 more answers
12. An organ pipe that is 1.75 m long and open at both ends produces sound of
podryga [215]

Answer:

354 m/s

Explanation:

For the second overtune (Third harmonic) of an open pipe,

λ = 2L/3................................ Equation 1

Where L = Length of the open pipe, λ = Wave length.

Given: L = 1.75 m.

Substitute into equation 1

λ = 2(1.75)/3

λ = 1.17 m.

From the question,

V = λf.......................... Equation 2

V = speed of sound in the room, f = frequency

Given: f = 303 Hz.

Substitute into equation 2

V = 1.17(303)

V = 353.5

V ≈ 354 m/s

Hence the right answer is 354 m/s

8 0
3 years ago
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