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Vilka [71]
3 years ago
6

The _______ is responsible for determining the frequency of vibration of the air column in the tube within a wind instrument. A.

effective length of the tube B. reed's frequency of vibration C. humidity of the air D. type of reed used
Physics
1 answer:
Vladimir [108]3 years ago
8 0

"A is correct answer." The effective length of the tube is responsible for determining the frequency of vibration of the air column in the tube within a wind instrument. "Hope this helps!" "Have a great day!" "Thank you for posting your question!"

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A crate of 45.2-kg tools rests on a horizontal floor. You exert a gradually increasing horizontal push on it and observe that th
iragen [17]

Answer:

 μ = 0.725

Explanation:

This problem refers to Newton's second law.

       F = ma

Let's write the equations on each axis

Y Axis

      N-W = 0

     N = W

    N = mg

X axis

   F-fr = ma

With the body not started moving its acceleration is zero

  F-fr = 0

  F = fr

The friction force equation is

  fr = μ N

  fr = μ m g

Let's replace and calculate

   F = μ m g

   μ = F / mg

   μ = 321 /45.2 9.8

   μ = 0.725

8 0
3 years ago
A 6.0-kg box slides down an inclined plane that makes an angle of 39° with the horizontal. If the coefficient of kinetic frictio
jeka57 [31]

Answer:

a. 3.1 m/s^2

Explanation:

The equation of the forces along the directions parallel and perpendicular to the slope are:

- Along the parallel direction:

mg sin \theta - \mu_k R = ma

where :

m = 6.0 kg is the mass  of the box

g = 9.8 m/s^2 the acceleration of gravity

\theta=39^{\circ}  is the angle of the slope

\mu_k = 0.40 is the coefficient of friction

R is the normal reaction  

a is the acceleration

- Along the perpendicular direction:

R-mg cos \theta =0

From the 2nd equation, we get an expression for the reaction force:

R=mg cos \theta

And substituting into the 1st equation, we can find the acceleration:

mg sin \theta - \mu_k mg cos \theta = ma

Solving for a,

a=g sin \theta - \mu_k g cos \theta =(9.8)(sin 39^{\circ})-(0.40)(9.8)(cos 39^{\circ})=3.1 m/s^2

6 0
3 years ago
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