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Elodia [21]
3 years ago
7

a sound wave is determined to have a frequency of 1,000 hz and wavelength of 35cm. what is the speed of this wave?

Physics
1 answer:
natulia [17]3 years ago
3 0

Answer:

The speed of the sound wave is 350 ms^{-1}

Explanation:

Given

Frequency of sound wave = 1000Hz = 1000s^{-1} (since 1 s^{-1} = 1 Hertz (Hz))

Wavelength of sound wave = 35 cm

Speed of sound wave = ?

All waves have same the relation among frequency, sound and speed which is given by

v=fλ , where

v is the speed

f is frequency

λ is wavelength

Therefore, the speed of sound wave = 1000s^{-1} x 35 cm

                                                              = 35000 cms^{-1}

                                                              = 350 ms^{-1}

Hence the speed of the sound wave is 350 ms^{-1}

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Alcohol of mass 33.2g and density 0.79kg/m³ or 790kg/m³ is mixed with water of 9g. What is the density of the resulting mixture?
KATRIN_1 [288]

Answer:

  0.83 g/cm³

Explanation:

The volume of the alcohol is ...

  (33.2 g)/(0.79 g/cm³) ≈ 42.0253 cm³

The density of water is about 1 g/cm³, so the volume of 9 g of it is ...

  (9 g)/(1 g/cm³) = 9 cm³

The total volume is ...

  42.0253 cm³ +9 cm³ = 51.0253 cm³

The total mass is ...

  33.2 g + 9 g = 42.4 g

So, the resulting density is ...

  (42.4 g)/(51.0253 cm³) ≈ 0.83 g/cm³

The resulting mixture has a density of about 0.83 g/cm³.

_____

<em>Additional comment</em>

Alcohol dissolves in water, so the total volume will be slightly less than 51.0253 cm³. The attached curve shows the result of mixing ethanol and water.

The weight of a mole of ethanol is about 46 g, of water, about 18.02 g. Then the mole fraction of alcohol is ...

  (33.2/46)/(33.2/46 +9/18.02) ≈ 0.59

The volume of the mix is then estimated to be (-1.05 cc/mol)(1.221 mol), or about 1.28 cm³ less than the volume indicated above. That brings the density up to about 0.85 g/cm³.

We're not completely sure of the relevance of this calculation, since many of the applicable parameters are not specified. The point is that <em>the density of the mix will probably be slightly more than the value calculated above</em>. YMMV

3 0
3 years ago
The technology in which parabolic mirror arrays are placed in a desert with a water-filled pipe traversing the array is used to
andriy [413]

Answer:

The sun

Explanation:

In this system the energy of the sun heats the water in the pipe, producing a high pressured steam, which is used for moving a turbine and producing electricity, is a transformation of energy from solar to thermal, then to mechanical to electrical.

4 0
3 years ago
Can you travel faster by not running forward ?
GrogVix [38]
Yeah i think with a car or a plane:)
5 0
2 years ago
A ball is shot from the ground into the air. At a height of 8.8 m, the velocity is observed to be
Mariulka [41]

Answer:

h = 10.4 m

R = 22.48 m

v= 16,2 m/s , α = 61.7°, below the horizontal

v = (7.7)i + (-14.3)j in meters per second (i horizontal, j downward)

Explanation:

The ball describes a parabolic path, and the equations of the movement are:

Equation of the uniform rectilinear motion (horizontal ) :

x = vx*t  :

Equations of the uniformly accelerated rectilinear motion of upward   (vertical ).

y = (v₀y)*t - (1/2)*g*t² Equation (2)

vfy² = v₀y² -2gy Equation (3)

vfy = v₀y -gt Equation (4)

Where:  

x: horizontal position in meters (m)

t : time (s)

vx: horizontal velocity  in m  

y: vertical position in meters (m)  

v₀y: initial  vertical velocity  in m/s  

vfy: final  vertical velocity  in m/s  

g: acceleration due to gravity in m/s²

Known data

y= 8.8 m

v = ( (7.7)i + (5.7)j  ) m/s : vx= 7.7 m/s , vy= 5.7 m/s

g = 9.8 m/s²

Calculation of the  initial  vertical velocity ( v₀y)

We apply Equation (3) with the known data

(vfy)² = (v₀y)² -2*g*y

(5.7)² = (v₀y)²- (2)*(9.8)*(8.8)

(5.7)²+ 172.48 =  (v₀y)²

v_{oy} = \sqrt{(5.7)^{2}+ 172.48 }

v₀y = 14.3 m/s

Calculation of the maximum height  the ball rise (h)

In the maximum height vfy=0

We apply the Equation (3) :

(vfy)² = (v₀y)² -2*g*y

0 = (14.3)² - 2*98*h

h = (14.3)² / 19.6

h = 10.4 m

Calculation of the time it takes for the ball to the maximum height

We apply the Equation (4) :

vfy = v₀y -gt

0 = v₀y -gt

gt = v₀y

t = v₀y/g

t = 14.3/9.8

t= 1.46 s

Flight time = 2t = 2.92 s

Total horizontal distance traveled by the ball  (R)

We replace data in the equation (1)

x =vx*t    vx= 7.7 m/s , t =2.92 s  (Flight time)

R = (7.7)* (2.92) = 22.48 m

Velocity of the ball (magnitude (v) and direction (α)) the instant before it hits the ground

vx = 7.7 m/s

vy = v₀y -gt = 14.3 - 9.8* (2.92) = -14.3 m/s

v= \sqrt{v_{x}^{2}+v_{y}^{2}  }

v= \sqrt{(7.7)^{2}+ (-14.3)^{2}  }

v= 16,2 m/s

\alpha = tan^{-1} (\frac{v_{y} }{v_{x} })

\alpha = tan^{-1} (\frac{-14.3 }{7.7 })

α = -61.7°

α = 61.7°, below the horizontal

i- j components of the v

v = (7.7)i + (-14.3)j in meters per second (i horizontal, j downward)

5 0
3 years ago
A horizontal pipe contains water at a pressure of 110 kPa flowing with a speed of 1.4 m/s. When the pipe narrows to one half its
Pavel [41]

Answer:

a

  v_2 =  5.6 \  m/s

b

   P_2 = 80600 \  Pa

Explanation:

From the question we are told that  

     The pressure of the water in the pipe is  P_1= 110 \  kPa  =  110 *10^{3 } \  Pa

      The speed of the water  is v_1 =  1.4 \  m/s

       The original area of the pipe is  A_1 =  \pi \frac{d^2 }{4}

       The  new area of the pipe is  A_2 = \pi *  \frac{[\frac{d}{2} ]^2}{4}  =  \pi *  \frac{\frac{d^2}{4} }{4} = \pi \frac{d^2}{16}

         

Generally the continuity equation is mathematically represented as

       A_1 *  v_1 =  A_2 * v_2

Here v_2 is the new velocity  

So

        \pi * \frac{d^2}{4}   *  1.4  = \pi * \frac{d^2}{16}   * v_2

=>     \frac{d^2}{4}   *  1.4  =  \frac{d^2}{16}   * v_2

=>    d^2    *  1.4  =  \frac{d^2}{4}   * v_2

=>    1.4  = 0.25    * v_2

=>     v_2 =  5.6 \  m/s

Generally given that the height of the original pipe and the narrower pipe are the same , then we will b making use of the  Bernoulli's equation for constant height to calculate the pressure

This is mathematically represented as

       

             P_1 + \frac{1}{2}  *  \rho *  v_1 ^2  =  P_2 + \frac{1}{2}  *  \rho *  v_2 ^2

Here \rho is the density of water with value  \rho =  1000  \  kg /m^3

             P_2 =  P_1 + \frac{1}{2} *  \rho [ v_1^2 - v_2^2 ]

=>          P_2 =  110 *10^{3} + \frac{1}{2} *  1000 *  [ 1.4 ^2 - 5.6 ^2 ]

=>          P_2 = 80600 \  Pa

4 0
2 years ago
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