Answer:
Sound wave types - longitudinal waves
Longitudinal waves - Vibrating string the creates sound in the way it moves.
Explanation:
Longitudinal waves have particles of the medium that are displaced in a parallel direction to energy transport.
Given that force is applied at an angle of 30 degree below the horizontal
So let say force applied if F
now its two components are given as
![F_x = Fcos30](https://tex.z-dn.net/?f=F_x%20%3D%20Fcos30)
![F_y = Fsin30](https://tex.z-dn.net/?f=F_y%20%3D%20Fsin30)
Now the normal force on the block is given as
![N = Fsin30 + mg](https://tex.z-dn.net/?f=N%20%3D%20Fsin30%20%2B%20mg)
![N = 0.5F + (18\times 9.8)](https://tex.z-dn.net/?f=N%20%3D%200.5F%20%2B%20%2818%5Ctimes%209.8%29)
![N = 0.5F + 176.4](https://tex.z-dn.net/?f=N%20%3D%200.5F%20%2B%20176.4)
now the friction force on the cart is given as
![F_f = \mu N](https://tex.z-dn.net/?f=F_f%20%3D%20%5Cmu%20N)
![F_f = 0.625(0.5F + 176.4)](https://tex.z-dn.net/?f=F_f%20%3D%200.625%280.5F%20%2B%20176.4%29)
![F_f = 110.25 + 0.3125F](https://tex.z-dn.net/?f=F_f%20%3D%20110.25%20%2B%200.3125F)
now if cart moves with constant speed then net force on cart must be zero
so now we have
![F_f + F_x = 0](https://tex.z-dn.net/?f=F_f%20%2B%20F_x%20%3D%200)
![Fcos30 - (110.25 + 0.3125F) = 0](https://tex.z-dn.net/?f=Fcos30%20-%20%28110.25%20%2B%200.3125F%29%20%3D%200)
![0.866F - 0.3125F = 110.25](https://tex.z-dn.net/?f=0.866F%20-%200.3125F%20%3D%20110.25)
![F = 199.2 N](https://tex.z-dn.net/?f=F%20%3D%20199.2%20N)
so the force must be 199.2 N
Mercury has a high boiling point of 357 degrees C.
Mercury has a freezing point of −39 degrees C.
Answer:
The energy that the truck lose to air resistance per hour is 87.47MJ
Explanation:
To solve this exercise it is necessary to compile the concepts of kinetic energy because of the drag force given in aerodynamic bodies. According to the theory we know that the drag force is defined by
![F_D=\frac{1}{2}\rhoC_dAV^2](https://tex.z-dn.net/?f=F_D%3D%5Cfrac%7B1%7D%7B2%7D%5CrhoC_dAV%5E2)
Our values are:
![V=30.1m/s](https://tex.z-dn.net/?f=V%3D30.1m%2Fs)
![C_d=0.45](https://tex.z-dn.net/?f=C_d%3D0.45)
![A=3.3m^2](https://tex.z-dn.net/?f=A%3D3.3m%5E2)
![\rho=1.2kg/m^3](https://tex.z-dn.net/?f=%5Crho%3D1.2kg%2Fm%5E3)
Replacing,
![F_D=\frac{1}{2}(1.2)(0.45)(3.3)(30.1)^2](https://tex.z-dn.net/?f=F_D%3D%5Cfrac%7B1%7D%7B2%7D%281.2%29%280.45%29%283.3%29%2830.1%29%5E2)
![F_D=807.25N](https://tex.z-dn.net/?f=F_D%3D807.25N)
We need calculate now the energy lost through a time T, then,
![W = F_D d](https://tex.z-dn.net/?f=W%20%3D%20F_D%20d)
But we know that d is equal to
![d=vt](https://tex.z-dn.net/?f=d%3Dvt)
Where
v is the velocity and t the time. However the time is given in seconds but for this problem we need the time in hours, so,
![W=(807.25N)(30.1m/s)(3600s/1hr)](https://tex.z-dn.net/?f=W%3D%28807.25N%29%2830.1m%2Fs%29%283600s%2F1hr%29)
(per hour)
Therefore the energy that the truck lose to air resistance per hour is 87.47MJ