Answer
given,
height of object = 2.7 cm
distance left of lens (u₁)= 20 cm
focal length of lens(f₁)= 10 cm
the distance of image



v₁ = 20 cm
magnification of first lens


m₁ = -1
distance of object from the second lens
u₂ = 52-20 = 32 cm
f₂ = 48 cm
now,



v₁ = 624 cm
magnification of first lens


m₁ = -12
total magnification
m = m₁ m₂
m = (-1)(-12)
m = 12
height of image


h' = -32.4 cm
a) distance between image and second lens is equal to 624 cm
b) height of image is equal to 32.4 cm
Answer:
The star is at a distance of 100 parsecs.
Explanation:
The distance can be determined by means of the distance modulus:
(1)
Where M is the absolute magnitude, m is the apparent magnitude and d is the distance in units of parsec.
Therefore, d can be isolated from equation 1

Then, Applying logarithmic properties it is gotten:
(2)
The absolute magnitude is the intrinsic brightness of a star, while the apparent magnitude is the apparent brightness that a star will appear to have as is seen from the Earth.
Since both have the same spectral type is absolute magnitude will be the same.
Finally, equation 2 can be used:
Hence, the star is at a distance of 100 parsecs.
Key term:
Parsec: Parallax of arc seconds
The answer is is 730.5 days rounded, so A. 730
The pressure in a liquid at a given depth is called the hydrostatic pressure. This can be calculated using the hydrostatic equation:
P = ρgh
P = (1000 kg/m³)(9.8 m/s²)(4 m)
<h2>
P = 39200 Pa</h2>