Answer:
-0.233 m left of diverging lens and ( 0.12 - 0.233 ) = -113 m left of conversing
and
0.023 m right of diverging lens
Explanation:
given data
focal length f2 = 14 cm = -0.14 m
Separation s = 12 cm = 0.12 m
focal length f1 = 21 cm = 0.21 m
distance u1 = 38 cm
to find out
final image be located and Where will the image
solution
we find find image location i.e v2
so by lens formula v1 is
1/f = 1/u + 1/v ...............1
v1 = 1/(1/f1 - 1/u1)
v1 = 1/( 1/0.21 - 1/0.38)
v1 = 0.47 m
and
u2 = s - v1
u2 = 0.12 - 0.47
u2 = -0.35
so from equation 1
v2 = 1/(1/f2 - 1/u2)
v2 = 1/(-1/0.14 + 1/0.35)
v2 = -0.233 m
so -0.233 m left of diverging lens and ( 0.12 - 0.233 ) = -113 m left of conversing
and
for Separation s = 45 cm = 0.45 m
v1 = 1/(1/f1 - 1/u1)
v1 =0.47 m
and
u2 = s - v1
u2 = 0.45 - 0.47 =- 0.02 m
so
v2 = 1/(1/f2 - 1/u2)
v2 = 1/(-1/0.14 + 1/0.02)
v2 = 0.023
so here 0.023 m right of diverging lens
Answer:
a) h = 14 m
b) h = 88 cm
c) f = 0.054 Hz
d) f = 0.13 Hz
Explanation:
a) T = 2π√(L/g)
L = T²g/4π²
L = (45/6)²(9.8) / 4π² = 13.963...
b) ½mv² = mgh
h = v²/2g
h = 4.15²/ (2(9.8)) = 0.87869
c) f = 1/T = 1 / (2π√(14 / 1.62)) = 0.0542
d) f = 6/45 = 0.13333...
The Sun emits electromagnetic radiation at many wavelengths across the EM spectrum.
Answer:
Explanation:
We can use the equation of speed in terms of distance. We know that the speed of light is constant value so we will have:
Now we know that 
I hope it helps you!
