Answer:
The focal length of eye piece is 6.52 cm.
Explanation:
Given that,
Angular Magnification of the microscope M = -46
the distance between the lens in microscope L= 16 cm
The focal length of objective f₀ = 1.5 cm
Normal near point N = 25 cm
Have to find focal length of eye piece f ₙ =?
The angular magnification is given by
M ≈ - (L-fₙ)N/f₀fₙ
Rearranging for fₙ
fₙ =L(1 - Mf₀/N)⁺¹
=18/2.76
fₙ = 6.52 cm
The focal length of eye piece is 6.52 cm.
Eight and I don’t know what else to say but for sure 8
Answer:2.541
Explanation:
Well , Potential Energy = mgh
m=mass = 82
g=acceleration of gravity=9.80m/s^2
h=what we are looking for
PE=mgh
PE/(mg) = h
Substitute in the values:
1970/(82 x 9.8) = h 2.541
Ω₀ = the initial angular velocity (from rest)
t = 0.9 s, time for a revolution
θ = 2π rad, the angular distance traveled
Let
α = the angular acceleration
ω = the final angular velocity
The angular rotation obeys the equation
(1/2)*(α rad/s²)*(0.9 s)² = (2π rad)
α = 15.514 rad/s²
The final angular velocity is
ω = (15.514 rad/s²)*(0.9 s) = 13.963 rad/s
If the thrower's arm is r meters long, the tangential velocity of release will be
v = 13.963r m/s
Answer: 13.963 rad/s
White dwarfs<span> are formed from the collapse of low mass </span>stars<span>, less than about 10 time the mass of the Sun. This </span>star<span> loses most of its mass in a wind, leaving behind a core that is less than 1.44 solar mass. On the other hand,</span>neutron stars<span> are formed in the catastrophic collapse of the core of a massive </span>star.
