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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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HELP ME PLEASE URGENT WILL MARK AND 5 STARS
Artist 52 [7]

Answer:

option (B) is the correct option.

please thanks me and follow me

7 0
3 years ago
A racecar accelerates from rest at 6.5 m/s2 for 4.1 s. How fast will it be going at the end of that time?
Pie

Answer:

The final velocity of the car is 26.65 m/s.

Explanation:

Given;

acceleration of the racecar, a = 6.5 m/s²

initial velocity of the car, u = 0

time of motion, t = 4.1 s

The final velocity of the car is given by;

v = u + at

where;

v is the final velocity of the car

suvstitute the givens

v = 0 + (6.5)(4.1)

v = 26.65 m/s.

Therefore, the final velocity of the car is 26.65 m/s.

6 0
2 years ago
a bicycle travels at a velocity of -2.33 m/s and has a displacement of -58.3 m. how much time did it take?
Alexandra [31]
We know that velocity is equal to the total displacement of an object over time.
velocity =  \frac{displacement}{time}
Deriving from that equation, we can say that:
t =  \frac{d}{v}
Okay, so here it goes:
t =  \frac{58.3m}{2.33 \frac{m}{s} } \\ t = 25.02s
The bicycle took 25.02 seconds to displace at 58.3 meters.
5 0
3 years ago
An infinitely long straight wire has a uniform linear charge density of Derive the 4. equation for the electric field a distance
marshall27 [118]

Answer:

E = \frac{\lambda}{2\pi \epsilon_0 r}

Explanation:

Let the linear charge density of the charged wire is given as

\frac{q}{L} = \lambda

here we can use Gauss law to find the electric field at a distance r from wire

so here we will assume a Gaussian surface of cylinder shape around the wire

so we have

\int E. dA = \frac{q}{\epsilon_0}

here we have

E \int dA = \frac{\lambda L}{\epsilon_0}

E. 2\pi r L = \frac{\lambda L}{\epsilon_0}

so we have

E = \frac{\lambda}{2\pi \epsilon_0 r}

4 0
3 years ago
A jet aircraft is traveling at 262 m/s in hor-
NeTakaya

Solution :

Speed of the air craft, $S_a$ = 262 m/s

Fuel burns at the rate of, $S_b$ = 3.92 kg/s

Rate at which the engine takes in air, $S_{air}$ = 85.9 kg/s

Speed of the exhaust gas that are ejected relative to the aircraft, $S_{exh}$ =921 m/s

Therefore, the upward thrust of the jet engine is given by

$F=S_{air}(S_{exh}-S_a)+(S_b \times S_{exh})$

F = 85.9(921 - 262) + (3.92 x 921)

   = 4862635.79 + 3610.32

   = $4.8 \times 10^6 \ N$

Therefore thrust of the jet engine is $4.8 \times 10^6 \ N$.

3 0
2 years ago
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