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Paha777 [63]
2 years ago
11

An older camera has a lens with a focal length of 60mm and uses 34-mm-wide film to record its images. Using this camera, a photo

grapher takes a picture of the Golden Gate Bridge that completely spans the width of the film. Now he wants to take a picture of the bridge using his digital camera with its 14-mm-wide CCD detector. What focal length should this camera's lens have for the image of the bridge to cover the entire detector?
Physics
1 answer:
lesya692 [45]2 years ago
6 0

Answer:

24.71 mm

Explanation:

Distance is proportional to focal length, so

d∝f

which means

\frac{d'_1}{d'_2}=\frac{f_1}{f_2}

Magnification of first lens

M_2=-\frac{d'_1}{d_1}

                   and

M_2=\frac{h'_1}{h_1}

Similarly, magnification of second lens

M_2=-\frac{d'_2}{d_1}

                   and

M_2=\frac{h'_2}{h_1}

From the above equations we get

\frac{M_1}{M_2}=\frac{d'_1}{d_2'}

                   and

\frac{M_1}{M_2}=\frac{h'_1}{h_2'}

which means,

\frac{d'_1}{d_2'}=\frac{h'_1}{h_2'}

and

\frac{d'_1}{d_2'}=\frac{f_1}{f_2}

So, we get

\frac{f_1}{f_2}=\frac{h'_1}{h_2'}\\\Rightarrow f_2=f_1\times\frac{h_2'}{h'_1}\\\Rightarrow f_2=60\times\frac{14}{34}=24.71\ mm

∴ Focal length should this camera's lens is 24.71 mm

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l = 2.4 m

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T cos θ − m g = 0

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w = √g / lcosθ

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the angular momentum  L = mr× v

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natta225 [31]
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1 year ago
A packing crate with mass 80.0 kg is at rest on a horizontal, frictionless surface. At t = 0 a net horizontal force in the +x-di
Nataly [62]

Answer:

Final speed of the crate is 15 m/s

Explanation:

As we know that constant force F = 80 N is applied on the object for t = 12 s

Now we can use definition of force to find the speed after t = 12 s

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