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Angelina_Jolie [31]
3 years ago
11

You have a 35X objective lens in place, and the numerical aperture of the objective lens is 0.75. The numerical aperture of the

condenser lens is 1.25. What is the total magnification if the ocular lens is 10X, with the eyepieces spaced 3 cm apart?
Physics
1 answer:
olasank [31]3 years ago
8 0

Answer:

350x

Explanation:

In a microscope the objective has higher magnification than the eyepiece so, this is a microscope

The magnification of a microscope is given by the product of the magnifications of the eyepiece and and the objective.

Objective lens magnification = 35x =m_o

Eyepiece magnification = 10x =m_e

Total magnification

M=m_o\times m_e\\\Rightarrow M=35\times 10\\\Rightarrow M=350

Total magnification is 350x

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I don’t know how to answer this question? Can anyone help?
VMariaS [17]

Answer:

F=ma

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According to the question,

F=3*F= 3F

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3 years ago
A particle executes simple harmonic motion with an amplitude of 2.18 cm.
Bogdan [553]

Answer:

The positive displacement from the midpoint of its motion at the speed equal one half of its maximum speed is 3.56 cm.

Explanation:

Maximum speed is  :

                          v (max) = Aω

Speed v at any displacement y is given by  

v^{2} = w^{2} (A^{2} - y^{2})   ........................................................  i

And,

               v = \frac{1}{2} v (max)  

          or,  2 × v = Aω     ....................................................   ii

Eliminating  ω from equations i and ii,

                       \frac{1}{4} A^{2}  w^{2}  =  w^{2}  ( A^{2}  - y^{2})

                     or, y^{2} =  (\frac{3}{4}) A^{2}  =(\frac{3}{4}) 2.18^{2}

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3 0
3 years ago
A monkey running through the jungle goes 0.198 km straight to the East, then turns 15.8° deflection from straight East toward th
inn [45]

Answer:

|\vec r|=339.82\ m

\theta=-6.67^o

Explanation:

<u>Displacement </u>

It's a vector magnitude that measures the space traveled by a particle between an initial and a final position. The total displacement can be obtained by adding the vectors of each individual displacement. In the case of two displacements:

\vec r=\vec r_1+\vec r_2

Given a vector as its polar coordinates (r,\theta), the corresponding rectangular coordinates are computed with

x=rcos\theta

y=rsin\theta

And the vector is expressed as

\vec z==

The monkey first makes a displacement given by (0.198 km,0°). The angle is 0 because it goes to the East, the zero-reference for angles. Thus the first displacement is

\vec r_1==\ km=\ m

The second move is (145 m , -15.8°). The angle is negative because it points South of East. The second displacement is

\vec r_2==\ m

The total displacement is

\vec r=\ m+\ m

\vec r=\ m

In (magnitude,angle) form:

|\vec r|=\sqrt{337.52^2+(-39.48)^2}=339.82\ m

\boxed{|\vec r|=339.82\ m}

\displaystyle tan\theta=\frac{-39.48}{337.52}=-0.1169

\boxed{\theta=-6.67^o}

5 0
3 years ago
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