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:
It can be concluded that the star is moving away from the observer.
Explanation:
Spectral lines will be shifted to the blue part of the spectrum if the source of the observed light is moving toward the observer, or to the red part of the spectrum when is moving away from the observer (that is known as the Doppler effect).
The wavelength at rest for this case is 434 nm and 410 nm (
, 
)
Since, 
(444nm) is greater than 
(434 nm) and (420nm) is greater than (410 nm), it can be concluded that the star is moving away from the observer
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Below are the choices:
only the x component
only the y component
both the x and y components
neither the x nor the y component
The answer is neither the x nor the y component
Answer:
x = 0.75801 = 75.801%
T_2 = 72..78 degree F
Explanation:
From superheated R 134 a properties table
At 200 lb/in^2 and 200 degree F
steady flow energy equation is givena s
At 90 lb/in2 Tsat = 72.78 degree F
hfg = 77.345 Btu/lbm
h = hf + x hfg
solving for x we get
x = 0.75801 = 75.801%
