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igor_vitrenko [27]
3 years ago
11

What is the wavelength of a wave that has a speed of 3 km/s and a frequency of 12 Hz? A. 36 km B. 3.6 km C. 0.25 km D. 4 km

Physics
1 answer:
butalik [34]3 years ago
3 0

Answer:

c. 0.25km

Explanation:

v=f x wavelength

3000 = 12 x wavelength

wavelength = 3000/12 = 250m

250m to km

To convert m to km, we divide by 1000

250/1000 =0.25km

wavelength = 0.25km

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     Using the Universal Gratitation Law, we have:

F= \frac{MmG}{d^2}  \\ MmG=2*10^{20}*(3.84*10^8)^2 \\ MmG=29.4912*10^36
 
     Again applying the formula in the new situation, comes:

F= \frac{MmG}{d^2} \\ F= \frac{29.4912*10^36}{(1.92*10^8)^2} \\ \boxed {F=8*10^{20}}

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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

     

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