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sammy [17]
3 years ago
14

A periodic transverse wave has an amplitude of 0.20 meter and a wavelength of 3.0 meters. If the frequency of this wave is 12 Hz

, what is its speed?
Physics
1 answer:
Aleksandr [31]3 years ago
7 0

The speed is 36 m/s

<u>Explanation:</u>

Speed of any wave is the measure of how fast the wave is propagating or travelling in medium. So, the speed of the wave can be calculated as the product of frequency with wavelength of that wave.

Thus, speed is directly proportional to the frequency and wavelength of a wave.

Speed = Frequency \times wavelength

<u>Given data:</u>

frequency of the wave is given as 12 Hz

wavelength is given as 3 m

Therefore, the speed will be

Speed = 12 × 3 = 36 m/s

Thus, the speed of the transverse wave is 36 m/s.

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Note that the scale bar under the photo is labeled 1 μm (micrometer). The scale bar works the same way as a scale on a map, wher
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Answer:

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8 0
3 years ago
In the aftermath of an intense earthquake, the earth as a whole "rings" with a period of 54 minutes.
lawyer [7]

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The frequency of a wave is defined as the number of cycles completed per second while the period refers to the time taken to complete a cycle. The frequency is the inverse of period.

So;

Period(T) = 54 minutes or 3240 seconds

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6 0
2 years ago
A planet is discovered orbiting around a star in the galaxy Andromeda at four times the distance from the star as Earth is from
Dmitriy789 [7]

Answer:

Tp/Te = 2

Therefore, the orbital period of the planet is twice that of the earth's orbital period.

Explanation:

The orbital period of a planet around a star can be expressed mathematically as;

T = 2π√(r^3)/(Gm)

Where;

r = radius of orbit

G = gravitational constant

m = mass of the star

Given;

Let R represent radius of earth orbit and r the radius of planet orbit,

Let M represent the mass of sun and m the mass of the star.

r = 4R

m = 16M

For earth;

Te = 2π√(R^3)/(GM)

For planet;

Tp = 2π√(r^3)/(Gm)

Substituting the given values;

Tp = 2π√((4R)^3)/(16GM) = 2π√(64R^3)/(16GM)

Tp = 2π√(4R^3)/(GM)

Tp = 2 × 2π√(R^3)/(GM)

So,

Tp/Te = (2 × 2π√(R^3)/(GM))/( 2π√(R^3)/(GM))

Tp/Te = 2

Therefore, the orbital period of the planet is twice that of the earth's orbital period.

5 0
3 years ago
Read 2 more answers
Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is
Ronch [10]

Complete Question

A gas gun uses high pressure gas tp accelerate projectile through the gun barrel.

If the acceleration of the projective is : a = c/s m/s​2

Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is s= 1.5m and the projectile is initially at rest. The projectile accelerates until it reaches the end of the barrel at s=3m. What is the value of the constant c such that the projectile leaves the barrel with velocity of 200m/s?

Answer:

The value of the constant is  c = 28853.78 \ m^2 /s^2

Explanation:

From the question we are told that

         The acceleration is  a =  \frac{c}{s}\   m/s^2

         The  initial position of the projectile is s= 1.5m

         The final position of the projectile is s_f =  3 \ m

          The velocity is  v = 200 \ m/s

     Generally  time  =  \frac{ds}{dv}

   and  acceleration is a =  \frac{v}{time }

so

            a = v  \frac{dv}{ds}

 =>        vdv  =  a ds

             vdv  = \frac{c}{s}  ds

integrating both sides

           \int\limits^a_b  vdv  = \int\limits^c_d \frac{c}{s}  ds

Now for the limit

          a =  200 m/s

             b = 0 m/s  

         c = s= 3 m

          d =s_f= 1.5 m

So we have  

           \int\limits^{200}_{0}  vdv  = \int\limits^{3}_{1.5} \frac{c}{s}  ds

              [\frac{v^2}{2} ] \left | 200} \atop {0}} \right.  = c [ln s]\left | 3} \atop {1.5}} \right.

            \frac{200^2}{2}  =  c ln[\frac{3}{1.5} ]

=>           c = \frac{20000}{0.69315}

              c = 28853.78 \ m^2 /s^2

     

5 0
3 years ago
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