The angular magnification of the microscope is 78.125.
<h3>What is angular magnification?</h3>
the ratio of the angle at the eye that an optical instrument's image occupies to the angle at the eye that an object occupies while not being viewed through an optical instrument is called as angular magnification.
Thus we can calculate it as:
- focal length of objective lens, fo = 1.6 cm
- focal length of the eyepiece, fe = 2.5 cm
- Distance between the lenses, L = 15 cm
- Least distance of distinct vision for normal eye, d = 25 cm
The formula for the magnification of the compound microscope is given by
M = - 78.125
Thus, the magnification of the compound microscope is 78.125.
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Answer:
The answer is:
VA= 33 m/S
VB= 27.8 m/s
VC= 50 m/s
VD=66.7 m/s
VE=35.7 m/s
So, the Legs from the highest velocity to the lower are:
D - C - E - A - B
The explanation:
we are going to use the velocity formula:
Velocity = distance (m) / time (s)
So,
A) when D = 18 Km = 18000 m
and t = 9 min = 9*60 =540 s
So V = 18000m/ 540 S
= 33 m/S
B) when D = 25 Km = 25000 m
and t = 15 min = 15 * 60 = 900 S
So V = 25000m / 900S
= 27.8 m/s
C) when D = 24 Km = 24000 m
and t = 8 min = 8 * 60 = 480 s
So V = 24000m / 480s
= 50 m/s
D) when D = 48 Km = 48000 m
and t = 12 min = 12 * 60 = 720 s
So V = 48000 m / 720s
=66.7 m/s
E) when D = 15 km = 15000m
and t = 7 min = 7 * 60 = 420 s
So V = 15000m / 420s
= 35.7 m/s
Answer:
0.211 kg
Explanation:
specific heat capacity of copper = 385 J/Kgk
heat loss by copper = mcθ = 3.8 × (8 - 84 ) × 385 = - 111188 J
heat needed to raise the temperature of water from 0°C to 8°C
= mcθ = 1.2 kg × 4180 × ( 8 - 0) = 40128 J
111188 J - 40128 J = 71060
71060 = ml
71060 / 336000 = mass of ice where latent heat of fusion = 3.36 × 10⁵JKg⁻¹
m = 0.211 kg
The correct answer should be time
- P is power
- R is resistance
Hence
- Therefore if power is low then resistance will be high.
The first bulb has less power hence it has greater filament resistance.