Answer:
The frequency of sound heard by the boy is 1181 Hz.
Explanation:
Given that,
Frequency of sound from alarm ![f_{0} = 1210\ Hz](https://tex.z-dn.net/?f=f_%7B0%7D%20%3D%201210%5C%20Hz)
Speed = -8.25 m/s
Negative sign show the boy riding away from the car
Speed of sound = 343
We need to calculate the heard frequency
Using formula of frequency
![f = f_{0}(\dfrac{v+v_{0}}{v-v_{s}})](https://tex.z-dn.net/?f=f%20%3D%20f_%7B0%7D%28%5Cdfrac%7Bv%2Bv_%7B0%7D%7D%7Bv-v_%7Bs%7D%7D%29)
Where,
= frequency of source
= speed of observer
= speed of source
= speed of sound
Put the value into the formula
![f=1210\times\dfrac{343+(-8.25)}{343-0}](https://tex.z-dn.net/?f=f%3D1210%5Ctimes%5Cdfrac%7B343%2B%28-8.25%29%7D%7B343-0%7D)
here, source is at rest
![f=1180.8\ Hz](https://tex.z-dn.net/?f=f%3D1180.8%5C%20Hz)
![f=1181\ Hz](https://tex.z-dn.net/?f=f%3D1181%5C%20Hz)
Hence, The frequency of sound heard by the boy is 1181 Hz.
Longggggg duhhhhhh !!!!!!!!
So I am not 100% on this but when I was learning about atoms I remember that the<span> charge: the amount of the charge has nothing to do with size. I think that electrons cancel out protons so that there is only one real Charge going</span>
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Answer:
![R_{max} = 2.125 R](https://tex.z-dn.net/?f=R_%7Bmax%7D%20%3D%202.125%20R)
Explanation:
If the maximum height attained by the rock is equal to the range of the rock
then we will say
![H = R](https://tex.z-dn.net/?f=H%20%3D%20R)
![\frac{v^2 sin^2\theta}{2g} = \frac{v^2 sin(2\theta)}{g}](https://tex.z-dn.net/?f=%5Cfrac%7Bv%5E2%20sin%5E2%5Ctheta%7D%7B2g%7D%20%3D%20%5Cfrac%7Bv%5E2%20sin%282%5Ctheta%29%7D%7Bg%7D)
so from this we can say
![\frac{sin^2\theta}{2} = 2sin\theta cos\theta](https://tex.z-dn.net/?f=%5Cfrac%7Bsin%5E2%5Ctheta%7D%7B2%7D%20%3D%202sin%5Ctheta%20cos%5Ctheta)
![tan\theta = 4](https://tex.z-dn.net/?f=tan%5Ctheta%20%3D%204)
![\theta = 75.96 degree](https://tex.z-dn.net/?f=%5Ctheta%20%3D%2075.96%20degree)
now original range is given as
![R = \frac{v^2 sin(2\theta)}{g}](https://tex.z-dn.net/?f=R%20%3D%20%5Cfrac%7Bv%5E2%20sin%282%5Ctheta%29%7D%7Bg%7D)
![R = \frac{v^2 sin(2\times 75.96)}{g}](https://tex.z-dn.net/?f=R%20%3D%20%5Cfrac%7Bv%5E2%20sin%282%5Ctimes%2075.96%29%7D%7Bg%7D)
![R = 0.47\frac{v^2}{g}](https://tex.z-dn.net/?f=R%20%3D%200.47%5Cfrac%7Bv%5E2%7D%7Bg%7D)
now we know that for maximum possible range we need to throw at 45 degree
![R_{max} = \frac{v^2 sin(2\times 45)}{g}](https://tex.z-dn.net/?f=R_%7Bmax%7D%20%3D%20%5Cfrac%7Bv%5E2%20sin%282%5Ctimes%2045%29%7D%7Bg%7D)
![R_{max} = \frac{v^2}{g}](https://tex.z-dn.net/?f=R_%7Bmax%7D%20%3D%20%5Cfrac%7Bv%5E2%7D%7Bg%7D)
![R_{max} = 2.125 R](https://tex.z-dn.net/?f=R_%7Bmax%7D%20%3D%202.125%20R)