You have a piece of film paper that is 3 in x 5 in. You fix it inside the back of a pinhole camera with a focal length of 5.5 in. You want to use it to take a picture of your team’s mascot – a giant guinea pig that just barely fits in a 4 ft. tall cube. The picture will be taken directly in front, from a stool that places the aperture 2 ft. above the ground. You have to determine how far away the camera must be from your mascot to get a good portrait that fills up the whole film paper, without cutting any part of him off. How far apart should the camera and the mascot be to take the portrait? Show your work.

Answers

Answer 1

Answer:

In order to take a portrait, the distance of the mascot from the camera should be approximately 7.33 feet

Explanation:

The size of the film paper = 3 in. × 5 in.

The focal length of the camera = 5.5 in.

The height and width of the guinea pig = 4 ft.

The height of the aperture above the ground = 2 ft.

Therefore, we have;

Magnification = Height of image/(Height of object)

Withe the 3 in. wide film, we have;

Magnification = 3 in./(4 ft.) = 3 in./(48 in.) = 0.0625

Magnification = Length of camera/(Distance of object from pin hole)  

∴ Length of camera/(Distance of object from pin hole) = 0.0625

Length of camera = Focal length of the camera = 5.5 in.

Therefore;

5.5 in./(Distance of object from pin hole) = 0.0625

Distance of object from pin hole = 5.5/0.0625 = 88 inches = 7.33 ft

Therefore, the camera should be approximately 7.33 ft. from the mascot to take a portrait.

Answer 2

In this exercise we have to use the magnification knowledge to calculate the distance that the photograph should be taken, thus we have to:

Distance of the mascot from the camera should be approximately 7.33 feet

To calculate the best distance to take the photo, we have that some information must be taken into account such as:

Size of the film paper: [tex](3)*(5) in[/tex] Focal length of the camera: [tex]5.5 in[/tex] Height and width of the guinea pig: [tex]4 ft[/tex] Height of the aperture above the ground: [tex]2 ft[/tex]

Therefore, we have that the formula of magnification is:

[tex]Magnification = Height \ of \ image/(Height \ of \ object)[/tex]

With the 3 in wide film, we have;

[tex]Magnification = 3 in/(4 ft) \\= 3 in/(48 in) = 0.0625 in[/tex]

Rewriting the magnification formula as:

[tex]Magnification = Length \ of \ camera/(Distance \ of \ object \ from \ pin \ hole)[/tex]  

Substituting the values ​​already known we have the equation will be matched as:

[tex]Length\ of \ camera/(Distance\ of\ object \ from \ pin \ hole) = 0.0625\\Length \ of \ camera = Focal \ length \ of \ the \ camera = 5.5 in.[/tex]

Therefore;

[tex]5.5 /(Distance \ of \ object\ from \ pin\ hole) = 0.0625 in\\Distance \ of \ object\ from \ pin \ hole = 5.5/0.0625\\ = 88 inches = 7.33 ft[/tex]

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Related Questions

A sign that has a white background with a
green panel with white lettering is a
a. general information sign
b. safety instruction sign
c. caution sign
d. danger sign

Answers

Answer:

A

Explanation:

general information sign

safety is orange

caution is yellow

danger is red

I got u you are going to pass your class easy

Help me for this question

Answers

The answer is definitely c, your correct:) I hope you have a good day!!

You are designing an airplane to carry liquid cargo that will slosh and move side to side in its container. This could make the plane unstable. What type of airfoil would you use for the wing? Why. Answer in a full sentence or more

Answers

I wold use a flat bottom airfoil because I feel I will mack it mor stabilizes than others

To create an effective study schedule, a student must
guess the amount of time needed for studying.
O write down all activities and commitments.
O allow the same amount of time each day for studying.
O budget periods of time to study daily, even if it is interrupted.

Answers

Answer:

B) write down all activities and commitments.

Explanation:

also i found the answer on quizlet hope this works:))

Answer:

write down all activities and commitments.

Explanation:

i had got the same question on my unit test and it was the answer it was right

for high-volume production runs, machining parts from solid material might not be the best choice of manufacturing operations because

Answers

Answer:

There are actually multiple types of processes a manufacturer uses, and those can be grouped into four main categories: casting and molding, machining, joining, and shearing and forming.

Explanation:

The displacement volume of an internal combustion engine is 3 liters. The processes within each cylinder of the engine are modeled as an air-standard Diesel cycle with a cutoff ratio of 2.5. The state of the air at the beginning of compression is fixed by P1=95kPa, T1=22oC, and V1 = 3.17 liters.
Determine the net work per cycle, in kJ, the power developed by the engine, in kW, and the thermal efficiency, if the cycle is executed 1000 times per min.

Answers

Answer:

1) The power developed by the engine is 14705.7739 kW

2) The thermal efficiency is approximately 61.5%

Explanation:

The given parameters are;

P₁ = 95 kPa

T₁ = 22°C

V₁ = 3.17 liters

The cutoff ratio = 2.5

Displacement volume = 3 liters

The number of times the cycle is executed per minute = 1000 times per minute

We have;

The displacement volume = V₁ - V₂ = 3 l

V₁ = 3.17 l

V₂ = 3 - 3.17 = 0.17 l

Compression ratio = V₁/V₂ = 3.17/0.17 ≈ 18.65

P₂/P₁ = P₂/(95 kPa) =  (V₁/V₂)^(k) = 18.65^1.4

P₂ = (95×18.65^(1.4)) ≈ 5710.5 kPa

T₂/T₁ = (V₁/V₂)^(k - 1)

T₂/(295 K)= (18.65)^(1.4 - 1)

T₂ = 295 * (18.65)^(1.4 - 1) = 950.81 K

The cutoff ratio = V₃/V₂ = 2.5

T₃ = T₂ × V₃/V₂  = 2.5 * 950.81 K = 2377.025 K

[tex]Q_{in}[/tex] = [tex]C_p[/tex]×(T₃ - T₂) = 1.006 × (2377.025 - 950.81) = 1,434.77 kJ/kg

T₄ = T₃ × (V₃/V₄)^(k-1) =

Therefore,

[tex]T_4 = T_3 \times \left (\dfrac{r_c}{r} \right )^{k - 1} = 2377.025 \times \left( \dfrac{2.5}{18.65} \right )^{1.4 - 1} \approx 1064 \ K[/tex]

T₄ ≈ 1064 K

[tex]Q_{out}[/tex] = [tex]-C_v \times (T_4 - T_1)[/tex]

[tex]C_v = C_p/k = 1.006/1.4 \approx 0.7186 \ kJ/kg[/tex]

∴ [tex]Q_{out}[/tex] = 0.7186×(1064 - 295) = 552.6034 kJ/kg

1) The net work = [tex]Q_{in}[/tex] - [tex]Q_{out}[/tex] = 1,434.77 kJ/kg - 552.6034 kJ/kg ≈ 882.17 kJ/kg

The number of cycle per minute = 1000 rpm

The number of cycle per minute = 1000 rpm/60 = 16.67 cycles per second

The power developed by the engine = The number of cycles per second × The net work of the engine

Therefore;

The power developed by the engine = 16.67 cycles/second  × 882.17 kJ/kg

The power developed by the engine = 14705.7739 kW

2) Efficiency, [tex]\eta _{th}[/tex], is given as follows;

[tex]\eta _{th} = \dfrac{Q_{in}-Q_{out}}{Q_{in}} \times 100 = 1 - \dfrac{Q_{out}}{Q_{in}} \times 100= 1 - \dfrac{552.6034}{1434.77}\times 100 \approx 61.5\%[/tex]

Therefore, the thermal efficiency ≈ 61.5%.

Use the diagram to determine total resistance. (Round the FINAL answer to one decimal place.) Note: Use the rated voltage and wattage to determine resistance. Note: Ra and Rb represent the resistance of the wire. The 120 volts of the motor and lamps is their rated voltage and not the actual voltage. I know the answer is 22.7 Ohms. I get very close, but I am not 100%.. I need to be 100%. If you can also: with the same diagram I need the total current as well! Thank you very much.

Answers

Answer:

22.6622 Ω ≈ 22.7 Ω5.29516 A ≈ 5.3 A

Explanation:

For an electric motor, 1 hp = 746 W, so 1/2 hp = 373 W. Then the total wattage of the load is ...

  Pl = 373 +100 +100 +75 = 648 W

The equivalent resistance is ...

  R = V^2/P = (120 V)^2/(648 W) = 22 2/9 Ω

Adding the wiring resistance, we get a circuit resistance of ...

  22.2222 +0.22 +0.22 = 22.6622 Ω . . . . total circuit resistance

Then the circuit current is ...

  120 V/(22.6622 Ω) ≈ 5.29516 A

Circuit resistance is about 22.7 Ω; circuit current is about 5.3 A.

_____

Exact values are 5099/225 ohms and 27000/5099 amps. Of course, these assume the motor and lamps are purely resistive, which they are not.

What is a specialized accreditation? A. evaluation of the quality of instruction B. evaluation of a particular program C. evaluation of students studying in an organization D. evaluation of recreational facilities in an organization

Answers

Answer:

B. evaluation of a particular program

Explanation:

Before students enrol into any given discipline, they should first ensure that their program of choice is well accredited. In a specific program, specialized accreditation has the function of telling would be students if the program meets up with academic standards in the field

Accreditation is a way of assessing faculty and curriculum quality of schools to make sure that they are up to academic standards and are also preparing students to future success in the field.

What are the common approximations made in the analysis of heat exchangers?

Answers

Answer: making sure that they are up to date

Explanation:

Explain how engineering and science-related?

Answers

Answer:

While scientists study how nature works, engineers create new things, such as products, websites, environments, and experiences. Because engineers and scientists have different objectives, they follow different processes in their work.

Explanation:

Solve the compound inequality. 3x − 4 > 5 or 1 − 2x ≥ 7

Answers

x ≤ −3 or x > 3
inequality form

how do you make coke for steel?

Answers

Can you be a bit more specific plz and that will let me identify the answer

parallel circuits???

Answers

PLEASE GIVE BRAINLIST

A parallel circuit has two or more paths for current to flow through. Voltage is the same across each component of the parallel circuit. The sum of the currents through each path is equal to the total current that flows from the source.

HOPE THIS HELPED

How can you safely lift and support a vehicle

Answers

Answer:

Place the jack under the part of the vehicle that it should contact when raised. If you're using jack stands, place them near the jack. If you place your jack incorrectly, you can injure your car. To find the proper place to position the jack for your particular vehicle, check your owner's manual.

1. Looking at the case study provided under the Companion Material section, what is the main problem that is addressed in this case study? Maintenance workers painted over the pipes. Engineers lost all of their work because of the flood. The water pipes broke and flooded the space center. There was no paperwork for the maintenance work done on the pipes.

Answers

Incomplete question. However, I provided information that could assist you in identifying the main problem or issue addressed in any case study.

Explanation:

First, note that a case study is simply a learning aid that allows one to learn from a real-life scenario.

To determine the main problems of a case study one needs to:

Read the case as many times as possible to become familiar with the message been expressed. For example, by highlighting or underlining the most important facts it can help you to discover the main problem or issue. Check for any facts provided in the case study, by so doing you can identify the most important problems.

Thus, by taking these few steps you may be able to determine the main problem in that case study.

What are the main causes of injuries when using forklifts?

Answers

The forklift overturning is a very common way of getting injured from a forklift. Overturning the forklift means it tips over onto it's side due to the operator turning it too fast.

12. You need to be at the lift controls whenever the lift is in motion A) True B)False

Answers

the answer to that would be a) true

What are materials engineers trying to discover when they study different materials? Whether or not materials corrode and oxidize how materials perform and deteriorate if certain materials have been used in the past and at which historical ages which metals conduct heat

Answers

Answer:

Material engineers study various materials to discover the reason and cause of its existence, the chemical properties, how long it has been there, and how it impacts human life.

Explanation:

Material engineer is an engineering discipline that focuses to improve human life by studying the environment and the various elements or materials it holds.

It uses the power of pure science to test and analysis its findings, documents its features and exposes them to the world. Materials like metals and other elemental forms were all tested by these engineers to determine its history and chemical and biological use.

Answer:

how materials perform and deteriorate

Explanation:

Two wooden members of uniform rectangular cross section are joined using a simple glued scarf splice. The maximum allowable shearing stress and maximum allowable normal stress in the glued splice is 50 MPa and 100 MPa, respectively. The cross-section area of the glued member is 400 mm2. (a) What should the value of the angle  be to achieve maximum load Fmax? (b) What is the magnitude of the maximum load Fmax?

Answers

Answer:

a). α =  26.57  

b). Maximum load is 50 .kN

Explanation:

a).

The normal force is given by

N = σ A cosec β

where, σ is the normal stress

            A is the cross sectional area

Similarly, shear force is given by

S= τ A cosec β

where, τ is the shearing stress

Now from the figure,

tan β = S/N

        = τ/σ

Therefore, [tex]$\beta = \tan^{-1}(2)$[/tex]  = 63.43

α = 90 - β = 26.57

b).

The normal force is given by

[tex]$N=(100\times 10^6)(400\times 10^{-6}) \text{ cosec}\ 63.43$[/tex]

[tex]$N=44.78\times 10^3$[/tex] N

We have

[tex]$\Sigma F_y=0$[/tex]

∴ N - F sin β = 0

⇒ F = N / sin β

      = [tex]$\frac{44.72\times 10^3}{\sin(63.43)} = 50\times 10^3 N$[/tex]

Similarly,

The shear force is given by

S = τ A cosec β

  = [tex]$(50\times 10^6)(400\times 10^{-6}) \text{ cosec}\ 63.43 = 22.36\times 10^3 N$[/tex]

[tex]$\Sigma F_x=0$[/tex]

∴ S - F cos β = 0

⇒ F = S / cos β

[tex]$\frac{22.36\times 10^3}{\cos(63.43)} = 49.99\times 10^3 N$[/tex]

Therefore, force is 50 kN.

The production process of rods from machine "A" yields specimen with the following specs. Mean: µ(LA)=20.00mm, STD: s(LA)=0.50mm. You need to purchase new machine to increase the production capacity and accuracy together. If the new machine "B" will produce half of the total rods, what is the STD, s(LB), that needs to achieve the total STD, s(LT)=0.4mm? Assume Corr(A,B)=0.4

Answers

Answer: the standard deviation STD of machine B is s (Lb) = 0.4557

Explanation:

from the given data, machine A and machine B produce half of the rods

Lt = 0.5La + 0.5Lb

so

s² (Lt) = 0.5²s²(La) + 0.5²s²(Lb) + 0.5²(2)Cov (La, Lb)

but Cov (La, Lb) = Corr(La, Lb) s(La) s(Lb) = 0.4s (La) s(Lb)

so we substitute

s²(Lt) = 0.25s² (La) + 0.25s² (Lb) + 0.4s (La) s(Lb)

0.4² = 0.25 (0.5²) + 0.25s² (Lb) + (0.5)0.4(0.5) s(Lb)

0.64 = 0.25 + s²(Lb) + 0.4s(Lb)

s²(Lb) + 0.4s(Lb) - 0.39 = 0

s(Lb) = { -0.4 ± √(0.16 + (4*0.39)) } / 2

s (Lb) = 0.4557

therefore the standard deviation STD of machine B is s (Lb) = 0.4557

While out on the International Space Station, an engineer was able to gather a sample of a new type of unidentified rock. What knowledge will the engineer use to predict the potential of this new material?

Answers

Answer:

The engineer will conduct a variety of tests, including chemical, mechanical, electrical, and physical examinations, to determine the potential of the new material.

Explanation:

They will need to test the material, this will also help to determine its malleability.

Hope this helps!

Which statement demonstrates the most scientific observation?

Answers

Where are the statements to choose from

A picture of a ____ is the label pictogram on a chemical that warms of a product's acute toxicity

Answers

Answer:skull and crossbones

Explanation:

A tensile test uses a test specimen that has a gage length of 50 mm and an area = 206 mm2. During the test, the specimen yields under a load of 97,944 N. The corresponding gage length = 50.2 mm - this is at the 0.2 percent yield point. The maximum load of 162,699 N is reached at a gage length = 63 mm. If fracture occurs at a gage length of 71 mm, determine the percent elongation in % - enter your answer as a whole number, not as a fraction.

Answers

Answer:

The percent elongation in the length of the specimen is 42%

Explanation:

Given that:

The gage length of the original test specimen  [tex]L_o[/tex] = 50 mm

The final gage length [tex]L_f[/tex] = 71 mm

The area = 206 mm²

maximum load  =  162,699 N

To determine the percent elongation in %, we use the formula:

[tex]\%EL = \dfrac{L_f-L_o}{L_o}\times 100[/tex]

[tex]\%EL = \dfrac{71 \ mm-50 \ mm}{50 \ mm}\times 100[/tex]

[tex]\%EL = \dfrac{21 mm}{50 \ mm}\times 100[/tex]

[tex]\%EL = 0.42 \times 100[/tex]

[tex]\mathbf{\%EL = 42 \%}[/tex]

The percent elongation in the length of the specimen is 42%

__________ use cleaning solutions that eventually become spent and must be disposed of properly


a sump pits
b car washers
c part washers
d tire cleaners

Answers

Answer:Part washers

Explanation:

I did it..

The only option that uses cleaning solutions that become spent and must be disposed properly is;

Option C; Parts Washers

To answer this question, let us examine each option;

Option A; Sump Pits; These are basins at the lowest part of specific areas designed to allow all water collected in that basin area to drain into it. This has no business with cleaning solution.

Option B; Car washers; These are like machine dryers used to clean the exterior and interior parts of motor vehicles.

Option C; Part Washers; These are mechanical devices that are designed for cleaning to remove debris, grime, oil dirt, paint, and other substances that could potentially cause contamination from parts in order to get them ready for assembling, packaging, or even coating. They are basically used to clean parts to get them ready for functional use.

Option D; Tire cleaner is just used to clean tire but is not something that becomes spent and should be disposed properly as it is more like a liquid.

The only correct option is Option C.

Read more at; https://brainly.com/question/25024807

What is the period if the clock frequency is 3.5 GHz?

Answers

Answer:

Period = 0.2857 nanoseconds

Explanation:

We are told that frequency = 3.5 GHz

This is simply 3.5 × 10^(9) Hz

Now, from wave equations, Period is given by the formula;

Period = 1/frequency

Thus;

Period = 1/(3.5 × 10^(9))

Period = 0.2857 × 10^(-9) seconds

From conversions, we can simplify the answer.

1 second = 10^(-9) nanoseconds

Thus, 0.2857 × 10^(-9) seconds = 0.2857 × 10^(-9) × 10^(-9) nanoseconds = 0.2857 nanoseconds

Which items are NOT found on a
door?*
5 points
Cladding
Moulding
Weatherstrip
Check Strap
Striker
All of the above
None of the above

Answers

Answer:

None of the above cause thats what i put

- The four leading causes of death in the
construction industry include electrical
incidents, struck-by incidents, caught-in or
caught-between incidents, and
a. vehicular incidents
b. falls
C. radiation exposure
d. chemical burns

Answers

My best guess is b but I honestly don’t know

The heat transfer rate per unit area in a thermal circuit is equivalent to what quantity in an electric circuit? A. voltage B. current C. resistance D. inductance

Answers

Answer:

A. voltage

Explanation:

The rate of heat transfer per unit area in a thermal circuit is equivalent to the voltage in an electrical circuit.

This is because, within an electrical circuit the voltage is supplied by a generator, which can be batteries, batteries, or others. The moment the electrical voltage acts within a circuit, it will consist of the gradient of electrical potential between two points, which will cause the electrons to move from the point with the highest concentration to the point with the lowest concentration (similar to the transfer of heat from the thermal circuit) generating the electric current.

Wheel grinders need be equipped with an

Answers

Answer:

wheel guard

Explanation:

to protect our hands and reduce spark

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