Answer: K: 0K, C: -273.15 degrees, F: -459.67 degrees
Explanation:
Answer:
1.12 m
0.08291 m
Explanation:
u = Upstream velocity = 0.4 m/s
Re = Reynold's number =
(turbulent)
= Viscosity of water = ![1.12\times 10^{-6}\ Pas](https://tex.z-dn.net/?f=1.12%5Ctimes%2010%5E%7B-6%7D%5C%20Pas)
Here the flow is turbulent so we have the relation
![Re_{xcr}=\frac{ux_{cr}}{\nu}\\\Rightarrow x_{cr}=\frac{Re_{xcr}\nu}{u}\\\Rightarrow x_{cr}=\frac{5\times 10^5\times 1.12\times 10^{-6}}{0.4}\\\Rightarrow x_{cr}=1.4\ m](https://tex.z-dn.net/?f=Re_%7Bxcr%7D%3D%5Cfrac%7Bux_%7Bcr%7D%7D%7B%5Cnu%7D%5C%5C%5CRightarrow%20x_%7Bcr%7D%3D%5Cfrac%7BRe_%7Bxcr%7D%5Cnu%7D%7Bu%7D%5C%5C%5CRightarrow%20x_%7Bcr%7D%3D%5Cfrac%7B5%5Ctimes%2010%5E5%5Ctimes%201.12%5Ctimes%2010%5E%7B-6%7D%7D%7B0.4%7D%5C%5C%5CRightarrow%20x_%7Bcr%7D%3D1.4%5C%20m)
The approximate location downstream from the leading edge where the boundary layer becomes turbulent is 1.4 m
Boundary layer thickness relation is given by
![\delta={\frac{\nu x}{u}}^{\frac{1}{5}}\\\Rightarrow \delta={\frac{1.12\times 10^{-6}\times 1.4}{0.4}}^{\frac{1}{5}}\\\Rightarrow \delta=0.08291\ m](https://tex.z-dn.net/?f=%5Cdelta%3D%7B%5Cfrac%7B%5Cnu%20x%7D%7Bu%7D%7D%5E%7B%5Cfrac%7B1%7D%7B5%7D%7D%5C%5C%5CRightarrow%20%5Cdelta%3D%7B%5Cfrac%7B1.12%5Ctimes%2010%5E%7B-6%7D%5Ctimes%201.4%7D%7B0.4%7D%7D%5E%7B%5Cfrac%7B1%7D%7B5%7D%7D%5C%5C%5CRightarrow%20%5Cdelta%3D0.08291%5C%20m)
The boundary layer thickness is 0.08291 m
Answer:
Part a: The force between the tires and the ground is 765 lbf.
Part b: The force between the pallet and the boulder is 1016 lbf.
Explanation:
From the free body diagram on the Truck, boulder and pallet level
is the acceleration of truck which is 1 ft/s2
is the mass of the truck which is 5000 lb or 155.28 slugs
is the mass of the boulder which is 1000 lb or 31.06 slugs
is the mass of the pallet which is 200 lb or 6.211 slugs
is the acceleration of the pallet or boulder which is 0.5 ft/s2
Applying Newton's 2nd Law of Motion on pulley in vertical direction
![2T-(m_a+m_b)g=(m_a+m_b)a_a\\2T-(37.271)(32.2)=(37.271)(0.5)\\T=609.32 lb_f](https://tex.z-dn.net/?f=2T-%28m_a%2Bm_b%29g%3D%28m_a%2Bm_b%29a_a%5C%5C2T-%2837.271%29%2832.2%29%3D%2837.271%29%280.5%29%5C%5CT%3D609.32%20lb_f)
Part a
Applying Newton's 2nd Law of Motion on Truck in horizontal direction
![F-T=m_Ta_T\\F=609.32+(155.28)(1)\\F=765 lb_f](https://tex.z-dn.net/?f=F-T%3Dm_Ta_T%5C%5CF%3D609.32%2B%28155.28%29%281%29%5C%5CF%3D765%20lb_f)
So the force between the tires and the ground is 765 lbf.
Part b
Applying Newton's 2nd Law of Motion on Boulder in vertical direction
![F_{AB}-m_bg=m_ba_b\\F_{AB}=m_b(g+a_b)\\F_{AB}=31.056(32.2+0.5)\\F_{AB}=1016 lb_f](https://tex.z-dn.net/?f=F_%7BAB%7D-m_bg%3Dm_ba_b%5C%5CF_%7BAB%7D%3Dm_b%28g%2Ba_b%29%5C%5CF_%7BAB%7D%3D31.056%2832.2%2B0.5%29%5C%5CF_%7BAB%7D%3D1016%20lb_f)
So the force between the pallet and the boulder is 1016 lbf.
Answer:
the average induced emf in the second winding is ![7.9168*10^{-5}V](https://tex.z-dn.net/?f=7.9168%2A10%5E%7B-5%7DV)
Explanation:
The magnetic field inside the first solenoid is given by,
![B= \mu_0NI](https://tex.z-dn.net/?f=B%3D%20%5Cmu_0NI)
Where
is the permeability of the free space
N is the number of turns of solenoid per unit length
I is the current in the solenoid
A is the cross-sectional area of the wire
Replacing we have,
![B= (4*\pi*10^{-7})(90/cm (\frac{100cm}{1m}))(0.350A)](https://tex.z-dn.net/?f=B%3D%20%284%2A%5Cpi%2A10%5E%7B-7%7D%29%2890%2Fcm%20%28%5Cfrac%7B100cm%7D%7B1m%7D%29%29%280.350A%29)
![B = 3.9584*10^{-3}T](https://tex.z-dn.net/?f=B%20%3D%203.9584%2A10%5E%7B-3%7DT)
Thus average emf induced in the second windigs is,
![\epsilon{avg}=\N\frac{dB}{dt}](https://tex.z-dn.net/?f=%5Cepsilon%7Bavg%7D%3D%5CN%5Cfrac%7BdB%7D%7Bdt%7D)
![\epsilon_{avg}=\frac{d}{dt}(AB)](https://tex.z-dn.net/?f=%5Cepsilon_%7Bavg%7D%3D%5Cfrac%7Bd%7D%7Bdt%7D%28AB%29)
![\epsilon_{avg}=A\frac{dB}{dt}](https://tex.z-dn.net/?f=%5Cepsilon_%7Bavg%7D%3DA%5Cfrac%7BdB%7D%7Bdt%7D)
![\epsilon_{avg}= A\frac{dB}{dt}](https://tex.z-dn.net/?f=%5Cepsilon_%7Bavg%7D%3D%20A%5Cfrac%7BdB%7D%7Bdt%7D)
![\epsilon_{avg}=(8*10^{-4})\frac{3.9584*10^{-3}}{0.04}](https://tex.z-dn.net/?f=%5Cepsilon_%7Bavg%7D%3D%288%2A10%5E%7B-4%7D%29%5Cfrac%7B3.9584%2A10%5E%7B-3%7D%7D%7B0.04%7D)
![\epsilon_{avg} = 7.9168*10^{-5}V](https://tex.z-dn.net/?f=%5Cepsilon_%7Bavg%7D%20%3D%207.9168%2A10%5E%7B-5%7DV)
Therefore the average induced emf in the second winding is ![7.9168*10^{-5}V](https://tex.z-dn.net/?f=7.9168%2A10%5E%7B-5%7DV)