I think the answer would be tensile, I’m sorry if it’s wrong
For a stationary siren on a firehouse is blaring at 81Hz. Assume the speed of sound to be 343m/s, the frequency perceived is mathematically given as'
F=81.721Hz
<h3>What is the
frequency perceived by a
firefighter racing toward the station at 11km/h?</h3>
Generally, the equation for the doppler effect is mathematically given as
![F'=\frac{vs+v}{vs}*f](https://tex.z-dn.net/?f=F%27%3D%5Cfrac%7Bvs%2Bv%7D%7Bvs%7D%2Af)
Therefore
F=81(343+3.05556)/343
F=81.721Hz
In conclusion, the frequency is
F=81.721Hz
Read more about frequency
brainly.com/question/24623209
Answer:
- de Broglie wavelength of the bullet is 4.56 x 10⁻³⁴ m
- The value of the wavelength shows that wave nature of matter is insignificant for the bullet because it is larger than particles.
Explanation:
Given;
mass of the bullet, m = 1.9 g = 0.0019 kg
velocity of the bullet, v = 765 m/s
de Broglie wavelength of the bullet is given by;
![\lambda = \frac{h}{mv}](https://tex.z-dn.net/?f=%5Clambda%20%3D%20%5Cfrac%7Bh%7D%7Bmv%7D)
where;
h is Planck's constant = 6.626 x 10⁻³⁴ J/s
λ is de Broglie wavelength of the bullet
![\lambda = \frac{h}{mv}\\\\ \lambda =\frac{(6.626*10^{-34})}{(0.0019)(765)}\\\\ \lambda =4.56 *10^{-34} \ m](https://tex.z-dn.net/?f=%5Clambda%20%3D%20%5Cfrac%7Bh%7D%7Bmv%7D%5C%5C%5C%5C%20%5Clambda%20%3D%5Cfrac%7B%286.626%2A10%5E%7B-34%7D%29%7D%7B%280.0019%29%28765%29%7D%5C%5C%5C%5C%20%20%5Clambda%20%3D4.56%20%2A10%5E%7B-34%7D%20%5C%20m)
Thus, this value of the wavelength shows that wave nature of matter is insignificant for the bullet because it is larger than particles.