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2 years ago

Compare and contrast camera obscura with what you know about modern digital photography, including cell phones.

Physics

2 answers:

Sonja [21]2 years ago

5 0

Answer:

It captures images but does not preserve them.

Ugo [173]2 years ago

5 0

A camera obscura is a darkened room. For making pictures there was a small hole in one wall which allowed an image of the outside scene to project onto the opposite wall, where an artist would draw the outlines of the scenery objects by tracing around the significant lines in the image.

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In an experiment, an object is released from rest from the top of a building. Its speed is measured as it reaches a point that i

NikAS [45]

Answer:

so the speed will increase by 1.44 times then the initial speed if the distance is increased to double

Explanation:

As we know that the air friction or resistance due to air is neglected then we can use the equation of kinematics here

$v_f^2 - v_i^2 = 2 a d$

since we released it from rest so we have

$v_i = 0$

so here we have

$v_f = \sqrt{2gd}$

now if the distance is double then we have

$v_f' = \sqrt{2g(2d)}$

now from above two equations we can say that

$v_f' = \sqrt2 v_f$

so the speed will increase by 1.44 times then the initial speed if the distance is increased to double

4 0

3 years ago

What is center of mass?

yuradex [85]

“a point representing the mean position of the matter in a body or system.”

3 0

3 years ago

A metal wire breaks when its tension reaches 100 newton. If the radius and length of the wire were both doubled then it would br

serg [7]

Answer:

200 N

Explanation:

Since Young's modulus for the metal, E = σ/ε where σ = stress = F/A where F = force on metal and A = cross-sectional area, and ε = strain = e/L where e = extension of metal = change in length and L = length of metal wire.

So, E = σ/ε = FL/eA

Now, since at break extension = e.

So making e subject of the formula, we have

e = FL/EA = FL/Eπr² where r = radius of metal wire

Now, when the radius and length are doubled, we have our extension as e' = F'L'/Eπr'² where F' = new force on metal wire, L' = new length = 2L and r' = new radius = 2r

So, e' = F'(2L)/Eπ(2r)²

e' = 2F'L/4Eπr²

e' = F'L/2Eπr²

Since at breakage, both extensions are the same, e = e'

So, FL/Eπr² = F'L/2Eπr²

F = F'/2

F' = 2F

Since F = 100 N,

F' = 2 × 100 N = 200 N

So, If the radius and length of the wire were both doubled then it would break when the tension reached 200 Newtons.

7 0

3 years ago

Question 1 of 10

blsea [12.9K]

lmoooooi9okkkkjghedjydthaksidhqelzyakx

5 0

3 years ago

Read 2 more answers

Rank the ten objects from loudest to softest.

frez [133]

Um what are the ten objects..?

5 0

3 years ago