Answer:
a) 5.197rev/s
b) Kf/Ki =2.28
Explanation:
a) Angular momentum of the system L = Iw
ButLi=Lf
Kiwi =Ifwf
wf = (Ii/If)will = (4.65/3.4)×3.8=5.197rev/s
b)Kinetic energy KE= 0.5Iw^2
Ki = 0.5Iiwi^2
Kf=0.5Ifwf^2
Kf/Ki = Ifwf/Iiwi
Kf/Ki = (4.65/3.4))(5.197/3.8)
Kf/Ki = 1.22(1.368)^2
Kf/Ki = 2.28
Answer:
Such limitations are given below.
Explanation:
- Each pn junction provides limited measurements of maximum forwarding current, highest possible inversion voltage as well as the maximum output level.
- If controlled within certain adsorption conditions, the pn junction could very well offer satisfying performance.
- In connector operation, the maximum inversion voltage seems to be of significant importance.
The second one if it’s on edge
Answer:
<h2>
6.36 cm</h2>
Explanation:
Using the formula to first get the image distance
1/f = 1/u+1/v
f = focal length of the lens
u = object distance
v = image distance
Given f = 16.0 cm, u = 24.8 cm
1/v = 1/16 - 1/24.8
1/v = 0.0625-0.04032
1/v = 0.02218
v = 1/0.02218
v = 45.09 cm
To get the image height, we will us the magnification formula.
Mag = v/u = Hi/H
Hi = image height = ?
H = object height = 3.50 cm
45.09/24.8 = Hi/3.50
Hi = (45.09*3.50)/24.8
Hi = 6.36 cm
The image height is 6.36 cm
