Answer:
Torque, ![\tau=34.6\ N.m](https://tex.z-dn.net/?f=%5Ctau%3D34.6%5C%20N.m)
Explanation:
It is given that,
Length of the wrench, l = 0.5 m
Force acting on the wrench, F = 80 N
The force is acting upward at an angle of 60.0° with respect to a line from the bolt through the end of the wrench. We need to find the torque is applied to the nut. We know that torque acting on an object is equal to the cross product of force and distance. It is given by :
![\tau=Fr\ sin\theta](https://tex.z-dn.net/?f=%5Ctau%3DFr%5C%20sin%5Ctheta)
![\tau=80\times 0.5\ sin(60)](https://tex.z-dn.net/?f=%5Ctau%3D80%5Ctimes%200.5%5C%20sin%2860%29)
![\tau=34.6\ N.m](https://tex.z-dn.net/?f=%5Ctau%3D34.6%5C%20N.m)
So, the torque is applied to the nut is 34.6 N.m. Hence, this is the required solution.
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Answer:
The possible range of wavelengths in air produced by the instrument is 7.62 m and 0.914 m respectively.
Explanation:
Given that,
The notes produced by a tuba range in frequency from approximately 45 Hz to 375 Hz.
The speed of sound in air is 343 m/s.
To find,
The wavelength range for the corresponding frequency.
Solution,
The speed of sound is given by the following relation as :
![v=f_1\lambda_1](https://tex.z-dn.net/?f=v%3Df_1%5Clambda_1)
Wavelength for f = 45 Hz is,
![\lambda_1=\dfrac{v}{f_1}](https://tex.z-dn.net/?f=%5Clambda_1%3D%5Cdfrac%7Bv%7D%7Bf_1%7D)
![\lambda_1=\dfrac{343}{45}=7.62\ m](https://tex.z-dn.net/?f=%5Clambda_1%3D%5Cdfrac%7B343%7D%7B45%7D%3D7.62%5C%20m)
Wavelength for f = 375 Hz is,
![\lambda_2=\dfrac{v}{f_2}](https://tex.z-dn.net/?f=%5Clambda_2%3D%5Cdfrac%7Bv%7D%7Bf_2%7D)
![\lambda_2=\dfrac{343}{375}=0.914\ m/s](https://tex.z-dn.net/?f=%5Clambda_2%3D%5Cdfrac%7B343%7D%7B375%7D%3D0.914%5C%20m%2Fs)
So, the possible range of wavelengths in air produced by the instrument is 7.62 m and 0.914 m respectively.
Refraction
the fact or phenomenon of light, radio waves, etc. being deflected in passing obliquely through the interface between one medium and another or through a medium of varying density.