Answer:
A.) 3605.6 N
B.) 33.7 degree
Explanation:
To find the result force acting on the wing of the airplane, we need to resolve the forces into x and y components
Resolving into x component :
Sum of forces = 3500 - 500 = 3000N
Resolving into y component:
Sum of forces = 2000N
Resultant force Fr = sqrt ( Fx^2 + Fy^2)
Fr = sqrt ( 3000^2 + 2000^2 )
Fr = sqrt ( 9000000 + 4000000 )
Fr = sqrt ( 13000000)
Fr = 3605.6 N
Therefore, resultant force acting on the wing is 3605.6 N
The direction of the vector will be:
Tan Ø = Fy / Fx
Substitute Fx and Fy into the formula
Tan Ø = 2000 / 3000
Tan Ø = 0.66666
Ø = tan^-1(0. 66666)
Ø = 33.7 degree.
Answer:
Yes, the energy is not simply the sum of the individual binding energies at each site, it is the product of energy at each binding site of hemoglobin.
Explanation:
Myoglobin and hemoglobin are two different cells. Myoglobin binds only one oxygen while the hemoglobin has the ability to binds four oxygen atoms at its four sides. Myoglobin present in muscle tissue only while hemoglobin is present in the whole body. Oxyhemoglobin is formed when oxygen binds with hemoglobin cell. This oxygen is take to all cells and energy is released due to the breakdown of glucose molecules with this oxygen.
Answer:
Explanation:
The inlet specific volume of air is given by:
The mass flow rates is expressed as:
The energy balance for the system can the be expresses in the rate form as:
![E_{in}-E_{out}=\bigtriangleup \dot E=0\\\\E_{in}=E_{out}\\\\\dot m(h_1+0.5V_1^2)=\dot W_{out}+\dot m(h_2+0.5V_2^2)+Q_{out}\\\\\dot W_{out}=\dot m(h_2-h_1+0.5(V_2^2-V_1^2))=-m({cp(T_2-t_1)+0.5(V_2^2-V_1^2)})\\\\\\\dot W_{out}=-(10.42lbm/s)[(0.25\frac{Btu}{lbm.\textdegree F})(300-900)\textdegree F+0.5((700ft/s)^2-(350ft/s)^2)(\frac{1\frac{Btu}{lbm}}{25037ft^2/s^2})]\\\\\\\\=1486.5\frac{Btu}{s}](https://tex.z-dn.net/?f=E_%7Bin%7D-E_%7Bout%7D%3D%5Cbigtriangleup%20%5Cdot%20E%3D0%5C%5C%5C%5CE_%7Bin%7D%3DE_%7Bout%7D%5C%5C%5C%5C%5Cdot%20m%28h_1%2B0.5V_1%5E2%29%3D%5Cdot%20W_%7Bout%7D%2B%5Cdot%20m%28h_2%2B0.5V_2%5E2%29%2BQ_%7Bout%7D%5C%5C%5C%5C%5Cdot%20W_%7Bout%7D%3D%5Cdot%20m%28h_2-h_1%2B0.5%28V_2%5E2-V_1%5E2%29%29%3D-m%28%7Bcp%28T_2-t_1%29%2B0.5%28V_2%5E2-V_1%5E2%29%7D%29%5C%5C%5C%5C%5C%5C%5Cdot%20W_%7Bout%7D%3D-%2810.42lbm%2Fs%29%5B%280.25%5Cfrac%7BBtu%7D%7Blbm.%5Ctextdegree%20F%7D%29%28300-900%29%5Ctextdegree%20F%2B0.5%28%28700ft%2Fs%29%5E2-%28350ft%2Fs%29%5E2%29%28%5Cfrac%7B1%5Cfrac%7BBtu%7D%7Blbm%7D%7D%7B25037ft%5E2%2Fs%5E2%7D%29%5D%5C%5C%5C%5C%5C%5C%5C%5C%3D1486.5%5Cfrac%7BBtu%7D%7Bs%7D)
Hence, the mass flow rate of the air is 1486.5Btu/s
