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motikmotik
4 years ago
15

If the wavelength of a 4.65 ✕ 102 Hz sound in freshwater is 3.45 m, what is the speed of sound in fresh water?

Physics
2 answers:
natima [27]4 years ago
8 0

Answer:

Explanation:

Given that, the wavelength is 3.45m

λ=3.45m

Frequency is 4.65×10^2Hz

F=4.65×10^2Hz

Speed is given as

V=Fλ

Then,

V=4.65×10^2× 3.45

V=1604.25m/s

The speed of sound in fresh water is 1604.25m/s

allochka39001 [22]4 years ago
3 0

Answer:

1604m/s

Explanation:

Speed (v) of a sound is expressed as a function of its wavelength(¶) and frequency (f). It is expressed as the product of the frequency of the sound and its wavelength. Mathematically,

Speed = frequency × wavelength

v = f¶

Given frequency = 4.65×10²Hz

¶ = 3.45m

v = 4.65×10²×3.45

v = 1604m/s

Therefore the wavelength of the sound in fresh water is 1604m/s

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g A 1.5 kg projectile is fired from the edge of a 20-m tall building with an initial velocity of v0 = 45.6 m/s directed 30° abov
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Answer:

a) t = 5.40 s

b) E_{p} = 684.3 J

c) E_{k} = 1559.52 J

d) mv = 74.6 kgm/s

Explanation:

<u>We have</u>:

v₀ : initial speed = 45.6 m/s

θ: 30 °

h: height = 20 m

m: projectile's mass = 1.5 kg  

     

a) The flight time of the projectile as it lands at Q (when it impacts on the ground in the parabolic motion) can be calculated using the following equation:          

-20 = v_{0}sin(\theta)*t - \frac{1}{2}gt^{2}   (1)

Where g is the gravity = 9.81 g/s²

By solving equation (1) for t, we have:

t = 5.40 s

b) The gravitational potential energy (E_{p}) of the projectile at point P (at the maximum height in the parabolic motion) is the following:  

E_{p} = mgh_{P}

Where h_{P}: is the height at the point P. This can be calculated using the following equation:

h_{P} = h + \frac{v_{0}^{2}(sin(\theta))^{2}}{2g} = 20 m + \frac{(45.6 m/s)^{2}(sin(30))^{2}}{2*9.81 m/s^{2}} = 46.5 m        

Now, the gravitational potential energy is:

E_{p} = mgh_{P} = 1.5 kg*9.81 m/s^{2}*46.5 m = 684.3 J  

c) The kinetic energy of the projectile at point R (the same height as the edge of the building in the parabolic motion) is:

E_{k} = \frac{m*v_{R}^{2}}{2}          

Where v_{R} is the velocity at the point R, which is:

v_{R} = -45.6 m/s      

Now, the kinetic energy is:

E_{k} = \frac{m*v_{R}^{2}}{2} = \frac{1.5 kg*(-45.6 m/s)^{2}}{2} = 1559.52 J          

d) The magnitude and direction of impulse the projectile impacts on the ground can be calculated using the equation:

F*t = m*v_{Q}

Where F: is the force

          

The velocity at the point Q is:

v_{Q} = \sqrt{(v_{q_{y}})^{2} + (v_{q_{x}}})^{2}} = \sqrt{(-30.17 m/s)^{2} + (39.49 m/s)^{2}} = 49.7 m/s      

Hence, the magnitude and direction of impulse is:                                              

F*t = m*v_{Q} = 1.5 kg*49.7 m/s = 74.6 kgm/s    

I hope it helps you!  

3 0
3 years ago
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