All objects in orbit must follow the path of an ellipse (one of Keplers laws)
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
Answer:
Explanation:
Using the magnification formula.
Magnification = Image distance(v)/object distance(u) = Image Height(H1)/Object Height(H2)
M = v/u = H1/H2
v/u = H1/H2...1
3) Given the radius of curvature of the concave lens R = 20cm
Focal length F = R/2
f = 20/2
f = 10cm
Object distance u = 5cm
Object height H2= 5cm
To get the image distance v, we will use the mirror formula
1/f = 1/u+1/v
1/v = 1/10-1/5
1/v = (1-2)/10
1/v =-1/10
v = -10cm
Using the magnification formula
(10)/5 = H1/5
10 = H1
H1 = 10cm
Image height of the peg is 10cm
4) If u = 15cm
1/v = 1/f-1/u
1/v = 1/10-1/15
1/v = 3-2/30
1/v = 1/30
v = 30cm
30/15 = H1/5
15H1 = 150
H1/= 10cm
5) if u = 20cm
1/v = 1/f-1/u
1/v = 1/10-1/20
1/v = 2-1/20
1/v = 1/20
v = 20cm
20/20 = H1/5
20H1 = 100
H1 = 5cm
6) If u = 30cm
1/v = 1/f-1/u
1/v = 1/10-1/30
1/v = 3-1/30
1/v = 2/30
v = 30/2 cm
v =>15cm
15/30 = Hi/5
30H1 = 75
H1 = 75/30
H1 = 2.5cm
Answer:
Centripetal Acceleration 18.75 m/s^2, Rotational Kinetic Energy 843.75 J
Explanation:
a Linear acceleration (we cant find tangential acceleration with the givens so we will find centripetal)
a= ω^2*r
ω= 300rev/min
convert into rev/s
300/60= 5rev/s
a= 18.75m/s^2
b) use Krot= 1/2 Iω^2
plug in gives
1/2(30*2.25)(25)= 843.75 J
Drag from her armas would slow her down if she was spinning at a fast speed
