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zavuch27 [327]
2 years ago
5

A wave is moving at the rate of 40 cm/s. Its wavelength is 5 cm. What is the frequency of the wave? INCLUDE THE CORRECT UNIT!

Physics
1 answer:
miskamm [114]2 years ago
8 0

Answer:

Solution given:

velocity=40cm/s

wave length=5cm

we have

frequency =velocity/wavelength=40/5=8hertz.

the frequency of the <u>wave</u><u> </u><u>i</u><u>s</u><u> </u><u>8</u><u> </u><u>h</u><u>e</u><u>r</u><u>t</u><u>z</u><u>.</u>

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When celery is placed in a glass of pure water the solution inside its cells is _____ compared to the water.?
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The correct answer to the question above is hypertonic. When celery is being placed in a glass of pure water, the solution inside its cells is going to appear as hypertonic compared to the water. This means that the cells inside has a higher concentration than outside.
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A 3.0-kg block moves up a 40o incline with constant speed under the action of a 26-N force acting up and parallel to the incline
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Instead of moving back and forth, a conical pendulum moves in a circle at constant speed as its string traces out a cone (see fi
tigry1 [53]

Answer:

a

The  radial acceleration is  a_c  = 0.9574 m/s^2

b

The horizontal Tension is  T_x  = 0.3294 i  \ N

The vertical Tension is  T_y  =3.3712 j   \ N

Explanation:

The diagram illustrating this is shown on the first uploaded

From the question we are told that

   The length of the string is  L =  10.7 \ cm  =  0.107 \ m

     The mass of the bob is  m = 0.344 \  kg

     The angle made  by the string is  \theta  =  5.58^o

The centripetal force acting on the bob is mathematically represented as

         F  =  \frac{mv^2}{r}

Now From the diagram we see that this force is equivalent to

     F  =  Tsin \theta where T is the tension on the rope  and v is the linear velocity  

     So

          Tsin \theta  =   \frac{mv^2}{r}

Now the downward normal force acting on the bob is  mathematically represented as

          Tcos \theta = mg

So

       \frac{Tsin \ttheta }{Tcos \theta }  =  \frac{\frac{mv^2}{r} }{mg}

=>    tan \theta  =  \frac{v^2}{rg}

=>   g tan \theta  = \frac{v^2}{r}

The centripetal acceleration which the same as the radial acceleration  of the bob is mathematically represented as

      a_c  =  \frac{v^2}{r}

=>  a_c  = gtan \theta

substituting values

     a_c  =  9.8  *  tan (5.58)

     a_c  = 0.9574 m/s^2

The horizontal component is mathematically represented as

     T_x  = Tsin \theta = ma_c

substituting value

   T_x  = 0.344 *  0.9574

    T_x  = 0.3294 \ N

The vertical component of  tension is  

    T_y  =  T \ cos \theta  = mg

substituting value

     T_ y  =  0.344 * 9.8

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The vector representation of the T in term is of the tension on the horizontal and the tension on the vertical is  

         

       T  = T_x i  + T_y  j

substituting value  

      T  = [(0.3294) i  + (3.3712)j ] \  N

         

3 0
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Vaselesa [24]

Answer:

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Explanation:

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