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OverLord2011 [107]
3 years ago
5

Given the following lens combination:

Physics
2 answers:
natulia [17]3 years ago
7 0

Given:

Lens.........diameter ...fl#

eyepiece...2cm............5

objective...40cm........15

focal length of eyepiece = 2*5 = 10cm

focal length of objective = 40*15 = 600cm

magnification = FL obj / FL eyp = 600/10 = 60x


pickupchik [31]3 years ago
6 0

magnification = (40*15)/(2*5) = 60X


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Reading glasses of what power are needed for a person whose near point is 125 cm , so that he can read a computer screen at 54 c
Burka [1]

Answer:

1.11 dioptre

Explanation:

d_{i} = Distance of the image = - (125 - 2) = - 123 cm

d_{o} = Distance of the object = 54 - 2 = 52 cm

f = Focal length of the lens

Using the equation

\frac{1}{d_{o}} + \frac{1}{d_{i}} = \frac{1}{f}

\frac{1}{52} + \frac{- 1}{123} = \frac{1}{f}

f = 90.1 cm

Power of the lens is given as

P = \frac{100}{f}

P = \frac{100}{90.1}

P = 1.11 Dioptre

3 0
3 years ago
A thin, uniform stick of mass M and length L is at rest on a flat, frictionless surface to which one end of it is pinned. A smal
labwork [276]

Answer:

a)  I = (\frac{M}{3} + \frac{4m}{9}) L²  ,   b)     w = (\frac{27 M}{18 m} + 2)⁻¹  Lv₀

Explanation:

a) The moment of inertia is a scalar that represents the inertia in circular motion, therefore it is an additive quantity.

The moment of inertia of a rod held at one end is

         I₁ = 1/3 M L²

The moment of inertia of the mass at y = L

        I₂ = m y²

 

The total inertia method

        I = I₁ + I₂

        I = \frac{1}{3} M L² + m (\frac{2}{3} L)²

        I = (\frac{M}{3} +\frac{4m}{9} ) L²

   

b) The conservation of angular momentum, where the system is formed by the masses and the bar, in such a way that all the forces during the collision are internal.    

Initial instant. Before the crash  

       L₀ = I₂ w₀  

angular and linear velocity are related  

       w₀ = y v₀  

      w₀ = \frac{2}{3}L v₀  

      L₀ = I₂ y v₀  

Final moment. After the crash  

      L_{f} = I w

 

how angular momentum is conserved  

      L₀ = L_{f}

      I₂ y v₀ = I w

substitute

      m (\frac{2L}{3})² (\frac{2L}{3} v₀ =  (\frac{M}{3} +\frac{4m}{9} ) L²  w

      \frac{6}{27}  m L³ v₀ = (\frac{M}{3} +\frac{4m}{9} ) L²  w

        \frac{6}{27}  m L v₀ = (\frac{M}{3} +\frac{4m}{9} )   w

        L v₀ = (\frac{27 M}{18 m} + 2)  w     

       w = (\frac{27 M}{18 m} + 2)⁻¹  Lv₀

 

6 0
2 years ago
Which of the following are places that you can work out
LUCKY_DIMON [66]

Answer:

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3 0
3 years ago
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bezimeni [28]

Answer:

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Explanation:

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radius of the space station is given as

r = (0.5) d

r = (0.5) (20.0)

r = 10 m

a = acceleration produced at outer rim = 9.80 m/s²

v = speed at which it rotates

acceleration is given as

a = \frac{v^{2}}{r}

9.80 = \frac{v^{2}}{10}

v = 9.89 m/s

4 0
3 years ago
Kirchhoff's loop rule for circuit analysis is an expression of which of the following? Conservation of charge Conservation of en
Irina18 [472]

Kirchhoff's circuit laws are two equalities that deal with the current and potential difference (commonly known as voltage) in the lumped element model of electrical circuits. They were first described in 1845 by German physicist Gustav Kirchhoff. This generalized the work of Georg Ohm and preceded the work of Maxwell.

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