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Paha777 [63]
3 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]3 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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Answer:

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3 years ago
A solid uniformly charged insulating sphere has uniform volume charge density p and radius R. Apply Gauss's law to determine an
RUDIKE [14]

Answer:

electric field E = (1 /3 e₀) ρ r

Explanation:

For the application of the law of Gauss we must build a surface with a simple symmetry, in this case we build a spherical surface within the charged sphere and analyze the amount of charge by this surface.

The charge within our surface is

 

     ρ = Q / V

     Q ’= ρ V '

The volume of the sphere is V = 4/3 π r³

     Q ’= ρ 4/3 π r³

The symmetry of the sphere gives us which field is perpendicular to the surface, so the integral is reduced to the value of the electric field by the area

      I E da = Q ’/ ε₀

      E A = E 4 πi r² = Q ’/ ε₀

      E = (1/4 π ε₀) Q ’/ r²

Now you relate the fraction of load Q ’with the total load, for this we use that the density is constant

     

      R = Q ’/ V’ = Q / V

How you want the solution depending on the density (ρ) and the inner radius  (r)

      Q ’= R V’

      Q ’= ρ 4/3 π r³

      E = (1 /4π ε₀) (1 /r²) ρ 4/3 π r³

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3 years ago
A loaded 500 kg sled is traveling on smooth horizontal snow at 5 m/s when it suddenly comes to a rough region. The region is 10
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Answer:

400 N

Explanation:

Change of Kinetic Energy to Friction Wok

∆KE = W

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250 x (25 - 9) = f x 10

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2 years ago
An inexperienced researcher runs an experiment and sets his alpha level at .40 because he can't wait to get his firstsignificant
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The smaller the value of α the more difficult it is to reject the null hypothesis. Therefore, choosing a low value for α can reduce the likelihood of Type I errors. The result here is that if the null hypothesis is false, it may be more difficult to reject using a lower value for α. The alpha value or statistical significance threshold is arbitrary. Which value to use depends on your field of study.

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4 0
1 year ago
A positive kaon (K+) has a rest mass of 494 MeV/c² , whereas a proton has a rest mass of 938 MeV/c². If a kaon has a total energ
vitfil [10]

Answer:

<em>0.85c </em>

Explanation:

Rest mass of Kaon M_{0K} = 494 MeV/c²

Rest mass of proton M_{0P}  = 938 MeV/c²

The rest energy is gotten by multiplying the rest mass by the square of the speed of light c²

for the kaon, rest energy E_{0K} = 494c² MeV

for the proton, rest energy E_{0P} = 938c² MeV

Recall that the rest energy, and the total energy are related by..

E = γE_{0}

which can be written in this case as

E_{K} = γE_{0K} ...... equ 1

where E = total energy of the kaon, and

E_{0} = rest energy of the kaon

γ = relativistic factor = \frac{1}{\sqrt{1 - \beta ^{2} } }

where \beta = \frac{v}{c}

But, it is stated that the total energy of the kaon is equal to the rest mass of the proton or its equivalent rest energy, therefore...

E_{K} = E_{0P} ......equ 2

where E_{K} is the total energy of the kaon, and

E_{0P} is the rest energy of the proton.

From E_{K} = E_{0P} = 938c²    

equ 1 becomes

938c² = γ494c²

γ = 938c²/494c² = 1.89

γ = \frac{1}{\sqrt{1 - \beta ^{2} } } = 1.89

1.89\sqrt{1 - \beta ^{2} } = 1

squaring both sides, we get

3.57( 1 - \beta^{2}) = 1

3.57 - 3.57\beta^{2} = 1

2.57 = 3.57\beta^{2}

\beta^{2} = 2.57/3.57 = 0.72

\beta = \sqrt{0.72} = 0.85

but, \beta = \frac{v}{c}

v/c = 0.85

v = <em>0.85c </em>

7 0
3 years ago
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