Answer:
a) 
= 3,375 cm
, b) f₀ = 77.625 cm
Explanation:
The magnification of a telescope is, to see at the far point of vision (infinity image)
m = - f₀ /
The length of the tube is
L = f₀ +
a) The focal length of the eyepiece
L = - m +
L = (1-m)
= L / (1-m)
Let's calculate
= 81.0 / (1 - (-23.0)
= 3,375 cm
b) the focal length of the target
f₀ = -m
f₀ = 23 (3.68)
f₀ = 77.625 cm
Answer:
Index of expansion: 4.93
Δu = -340.8 kJ/kg
q = 232.2 kJ/kg
Explanation:
The index of expansion is the relationship of pressures:
pi/pf
The ideal gas equation:
p1*v1/T1 = p2*v2/T2
p2 = p1*v1*T2/(T2*v2)
500 C = 773 K
20 C = 293 K
p2 = 35*0.1*773/(293*1.3) = 7.1 bar
The index of expansion then is 35/7.1 = 4.93
The variation of specific internal energy is:
Δu = Cv * Δt
Δu = 0.71 * (20 - 500) = -340.8 kJ/kg
The first law of thermodynamics
q = l + Δu
The work will be the expansion work
l = p2*v2 - p1*v1
35 bar = 3500000 Pa
7.1 bar = 710000 Pa
q = p2*v2 - p1*v1 + Δu
q = 710000*1.3 - 3500000*0.1 - 340800 = 232200 J/kg = 232.2 kJ/kg
The freezing point ..... :)
Answer:
maximum amplitude = 0.08 m
Explanation:
Given that
Time period T= 0.58 s
acceleration of gravity g= 9.8 m/s²
We know that time period of simple harmonic motion given as
T = 2π/ω
0.58 = 2π/ω
ω = 10.83rad/s
ω=angular frequency
Lets take amplitude = A
The maximum acceleration given as
a= ω² A
The maximum acceleration should be equal to g ,then block does not separate
a= ω² A
9.8 = 10.83² A
A = 0.08m
maximum amplitude = 0.08 m
