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sukhopar [10]
3 years ago
11

A small insect viewed through a convex lens is 1.5 cmcm from the lens and appears 2.5 times larger than its actual size. Part A

What is the focal length of the lens
Physics
1 answer:
Art [367]3 years ago
3 0

Answer:

The focal length of the lens is 2.5 cm

Explanation:

Use the two equations for thin lenses combined: the one for magnification (m), and the one that relates distances of object d_o, of image d_i, and focal length;

m=\frac{h_i}{h_o} =-\frac{d_i}{d_o} \\  \\\frac{1}{d_i} +\frac{1}{d_o} =\frac{1}{f}

Since we know the value of the magnification (m), we can write the image distance in terms of the object distance, and then use it to replace the image distance in the second equation:

m=-\frac{d_i}{d_o} \\2.5=-\frac{d_i}{d_o}\\d_i=-2.5\,d_o

then, solving for the focal distance knowing that the object distance is 1.5 cm:

\frac{1}{d_i} +\frac{1}{d_o} =\frac{1}{f}\\-\frac{1}{2.5\,d_o} +\frac{1}{d_o} =\frac{1}{f}\\(2.5\,d_o\,f)\,(-\frac{1}{2.5\,d_o} +\frac{1}{d_o}) =\frac{1}{f}\,(2.5\,d_o\,f)\\-f+2.5\,f=2.5\,d_o\\1.5\,f=2.5\,d_o\\f=\frac{2.5\,d_o}{1.5} \\f=\frac{2.5\,(1.5\,\,cm)}{1.5}\\f=2.5\,\,cm

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Answer:

<h3>The answer is 8.5 kg</h3>

Explanation:

The mass of the object can be found by using the formula

m =  \frac{f}{a}  \\

where

f is the force

a is the acceleration

So we have

m =  \frac{34}{4}  =  \frac{17}{2}  \\

We have the final answer as

<h3>8.5 kg</h3>

Hope this helps you

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At summer camp, the swimming course runs the length (L) of a small lake. To determine the length of the course, the camp counsel
sleet_krkn [62]

Answer:

47 m

Explanation:

Data obtained from the question include the following:

Length of dry leg 1 (L1) = 40 m

Length of dry leg 2 (L2) = 25 m

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The length of the swimming course can be obtained by using pythagoras theory as shown below:

L² = L1² + L2²

L² = 40² + 25²

L² = 1600 + 625

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8_murik_8 [283]

Answer:

The transverse component of acceleration is 26.32 m/s^2 where as radial the component of acceleration is 8.77 m/s^2

Explanation:

As per the given data

u=π/4 rad

ω=u'=2 rad/s

α=u''=4 rad/s

r=e^u

So the transverse component of acceleration are given as

a_{\theta}=(ru''+2r'u')\\

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r=e^u\\r=e^{\pi/4}\\r=2.1932 m

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a_{\theta}=(ru''+2r'u')\\a_{\theta}=(2.1932\times 4+2\times 4.3864 \times 2)\\a_{\theta}=26.32 m/s\\

The transverse component of acceleration is 26.32 m/s^2

The radial component is given as

a_r=r''-r\theta'^2

Here

r''=e^u.u'^2+e^u u''\\r''=2.1932 \times (2)^2+2.1932\times 4\\r''=17.5456 m

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a_r=r''-ru'^2\\a_r=17.5456-2.1932\times (2)^2\\a_r=8.7728 m/s^2

The radial component of acceleration is 8.77 m/s^2

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