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Sidana [21]
2 years ago
10

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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(a) The stick is supported by a sharp point at the middle. On the left side, a weight of 100 g is suspended at 40 cm from the mi
Jet001 [13]

Answer:

20cm

Explanation:

Hello!

remember that the condition for a body to be at rest is that the sum of its moments and its forces be zero,

To solve this problem you must draw the free body diagram of the stick (attached image) and sum up moments at point 0 (where the sharp is located), which results in the following equation

(100g)(40cm)=x(200g)

X=\frac{(100)(40)}{(200)} =20cm

6 0
3 years ago
Which of the following calls one might hear a refree make?
mestny [16]

Answer: D) All of the above

Explanation:

7 0
2 years ago
Read 2 more answers
On a certain planet, which is perfectly spherically symmetric, the free-fall acceleration has magnitude g = go at the north pole
ohaa [14]
The reason why there is a difference between free-fall acceleration is a centrifugal force.
I attached a diagram that shows how this force aligns with the force of gravity.
From the diagram we can see that:
F=F_g-F_{cf}=mg'-mw^2r'cos(\alpha)\\ ma=mg'-mw^2r'cos(\alpha)\\ a=g'-w^2rcos^2(\alpha)\\
Where g' is the free-fall acceleration when there is no centrifugal force, r is the radius of the planet, and w is angular frequency of planet's rotation. \alpha is the latitude.
We can calculate g' and wr^2 from the given conditions in the problem.
g(90)=g_0;\ g_0= g'-w^2rcos^2(90)\\
g_0=g'\\
g(0)=ag_0;\ ag_0=g_0-w^2rcos^2(0)\\
ag_0=g_0-w^2r\\
w^2r=g_0(a-1)

Our final equation is:
g=g_0-g_0(a-1)cos^2(\alpha)
Colatitude is:
\alpha_c=90^\circ-\alpha
The answer is:
g=g_0-g_0(a-1)cos^2(90-9)=g_0-g_0(a-1)sin^2(9)

5 0
3 years ago
How much force, in g cm/s2
scoray [572]
In this question force is measured in  g cm/s2 so we know that to get the answer we times g by cm/s2 
50 × 20 = 1000
3 0
3 years ago
Read 2 more answers
A 2 kg ball is dropped above the surface of Planet X. If the gravitational field strength at the surface of Planet X is 5 N/kg,
Trava [24]

Given data:

* The mass of the ball is 2 kg.

* The gravitational field strength at the surface of planet X is 5 N/kg.

Solution:

The weight of the ball on the planet X is,

W=ma

where m is the mass of ball, a is the gravitational field strength,

Substituting the known values,

\begin{gathered} W=2\times5 \\ W=10\text{ N} \end{gathered}

Thus, the weight of the ball on the surface of planet X is 10 N.

3 0
1 year ago
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