Answer:
Friction of the road on the motorcycle in the opposite direction
Explanation:
Khanacademy
In a string of length L, the wavelength of the n-th harmonic of the standing wave produced in the string is given by:
![\lambda=\frac{2}{n} L](https://tex.z-dn.net/?f=%20%5Clambda%3D%5Cfrac%7B2%7D%7Bn%7D%20L%20)
The length of the string in this problem is L=3.5 m, therefore the wavelength of the 1st harmonic of the standing wave is:
![\lambda=\frac{2}{1} \cdot 3.5 m=7.0 m](https://tex.z-dn.net/?f=%20%5Clambda%3D%5Cfrac%7B2%7D%7B1%7D%20%5Ccdot%203.5%20m%3D7.0%20m)
The wavelength of the 2nd harmonic is:
![\lambda=\frac{2}{2} \cdot 3.5 m=3.5 m](https://tex.z-dn.net/?f=%20%5Clambda%3D%5Cfrac%7B2%7D%7B2%7D%20%5Ccdot%203.5%20m%3D3.5%20m)
The wavelength of the 4th harmonic is:
![\lambda=\frac{2}{4} \cdot 3.5 m=1.75 m](https://tex.z-dn.net/?f=%20%5Clambda%3D%5Cfrac%7B2%7D%7B4%7D%20%5Ccdot%203.5%20m%3D1.75%20m)
It is not possible to find any integer n such that
, therefore the correct options are A, B and D.
Answer:
a.18.5 m/s
b.1.98 s
Explanation:
We are given that
![\theta=35^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%3D35%5E%7B%5Ccirc%7D)
a.Let
be the initial velocity of the ball.
Distance,x=30 m
Height,h=1.8 m
![v_x=v_0cos\theta=v_0cos35](https://tex.z-dn.net/?f=v_x%3Dv_0cos%5Ctheta%3Dv_0cos35)
![v_y=v_0sin\theta=v_0sin35](https://tex.z-dn.net/?f=v_y%3Dv_0sin%5Ctheta%3Dv_0sin35)
![x=v_0cos\theta\times t=v_0cos35\times t](https://tex.z-dn.net/?f=x%3Dv_0cos%5Ctheta%5Ctimes%20t%3Dv_0cos35%5Ctimes%20t)
![t=\frac{30}{v_0cos35}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B30%7D%7Bv_0cos35%7D)
![h=v_yt-\frac{1}{2}gt^2](https://tex.z-dn.net/?f=h%3Dv_yt-%5Cfrac%7B1%7D%7B2%7Dgt%5E2)
Substitute the values
![1.8=v_0sin35\frac{30}{v_0cos35}-\frac{1}{2}(9.8)(\frac{30}{v_0cso35})^2](https://tex.z-dn.net/?f=1.8%3Dv_0sin35%5Cfrac%7B30%7D%7Bv_0cos35%7D-%5Cfrac%7B1%7D%7B2%7D%289.8%29%28%5Cfrac%7B30%7D%7Bv_0cso35%7D%29%5E2)
![1.8=30tan35-\frac{6574.6}{v^2_0}](https://tex.z-dn.net/?f=1.8%3D30tan35-%5Cfrac%7B6574.6%7D%7Bv%5E2_0%7D)
![\frac{6574.6}{v^2_0}=21-1.8=19.2](https://tex.z-dn.net/?f=%5Cfrac%7B6574.6%7D%7Bv%5E2_0%7D%3D21-1.8%3D19.2)
![v^2_0=\frac{6574.6}{19.2}](https://tex.z-dn.net/?f=v%5E2_0%3D%5Cfrac%7B6574.6%7D%7B19.2%7D)
![v_0=\sqrt{\frac{6574.6}{19.2}}=18.5 m/s](https://tex.z-dn.net/?f=v_0%3D%5Csqrt%7B%5Cfrac%7B6574.6%7D%7B19.2%7D%7D%3D18.5%20m%2Fs)
Initial velocity of the ball=18.5 m/s
b.Substitute the value then we get
![t=\frac{30}{18.5cos35}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B30%7D%7B18.5cos35%7D)
t=1.98 s
Hence, the time for the ball to reach the target=1.98 s
Yes because of the smoke you are creating in the air