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Snezhnost [94]
3 years ago
11

The built in flash in a compact camera is usally capable of giving correct exsposure for distance up to how many meters?

Physics
1 answer:
Brut [27]3 years ago
8 0

Answer:

An on-camera flash is an indispensible accessory for many photographers; it provides additional light when conditions become too dark to handhold your camera comfortably, allows you to achieve more balanced exposures in daylight conditions, permits freezing of fast-moving subjects and can also be used to control or trigger other flash light sources. Additionally, a flash can be used as a highly effective creative tool to establish an aesthetic that elevates your imagery when lighting conditions are considered less than stellar. The benefits of an external on-camera flash far outweigh those provided by a built-in camera flash, while the only drawback is keeping an additional piece of equipment.

On-Camera Flash versus Off-Camera Flash versus In-Camera Flash

The term on-camera flash simply refers to a type of strobe light (flash) that can connect directly with your camera. While it is referred to as “on-camera” this does not require the flash to be physically mounted on your camera. On-camera flashes can, and often are, used off-camera. This differs from other strobe-light sources, such as studio pack strobes and monolights in that these types of strobes are not meant to be physically connected to your camera (except under rare and unusual circumstances involving convoluted methods of adaptation). Additionally, on-camera flashes usually have a self-contained power supply, although external power sources can sometimes be used to improve performance or battery life.

On-camera external flash also refers to the type of external flash that can be used on your camera, compared to a built-in flash that is integrated into many cameras. An on-camera external flash performs better than a built-in flash in almost every regard with the one exception that it is not built into your camera. The ability to take the flash off your camera results in a significantly greater number of lighting options; far more than simply providing a blast of flat light to the scene to facilitate an adequate exposure. It is often not desirable to have your flash pointed squarely at the scene at hand; more often than not you will want to bounce the flash light off other surfaces and point in other directions to control the look of your flash. When using an in-camera flash, you are forced to use the flash at the given angle from which it extends.

Most built-in flashes are also located near the camera lens, which can often result in the red-eye effect  when photographing subjects in dimly lit conditions. Red-eye occurs because pupils dilate in dim light, the built-in flash is aligned with the lens's optical axis, its beam enters the eye and reflects back at the camera from the retina at the rear of the eye, which is quite red. Being able to use an on-camera flash source off-camera, from a different angle, will help to eliminate the red-eye effect in your photographs of people.

Guide Numbers, Manual Usage, Controlling Flash Power and Sync Speeds

Before delving into the automatic technology that is contained within most contemporary flashes, it is best to understand how to manually control and grasp a flash’s power. This is directly related to having an understanding of exposure ratios—how shutter speeds and apertures affect and balance each other—even though auto-exposure metering is available and often utilized for determining the best exposure settings.

Explanation:

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Hernando builds a simple DC series circuit with a standard D-cell battery and uses an ammeter to determine that the total curren
HACTEHA [7]

Answer: I = 9.0 A

Explanation: <u>Ohm's</u> <u>Law</u> is defined as the relationship between voltage and current: in a circuit, the voltage accross a conductor is directly proportional to the current.

The constant of proportionality is Resistance, whose unit is ohm (Ω).

so, Ohm's Law is

V = R.I

It is asked for the current, so, rearraging:

I=\frac{V}{R}

When added two more batteries, the voltage of the circuit triples:

V_{final}=3V_{initial}

Since they are directly proportional, if voltage triples, so does current:

I_{final}=3I_{initial}

I_{final}=3(3.0)

I_{final}= 9.0

The new current, when driven by 3 batteries is 9.0 A.

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3 years ago
A 4 kg textbook sits on a desk. It is pushed horizontally with a 50 N applied force against a 15 N frictional force.
ikadub [295]

Answer:

b) No acceleration in the vertical

c) 35N

d) 35N

e) 8.75\ m/sec^2

Explanation:

a) The situation can be shown in the free body diagram shown in the figure below where F is the applied force, Fr is the friction force, W is the weight of the book and N is the normal force exerted vertically up from the desk to the book

b) The vertical movement is restrained by the normal force which opposes to the weight. In absence of any other force, they both are in equilibrium and the net force is zero

c) The net horizontal force acting on the book is the vectorial sum of the applied force and the friction force. Since they lie in the same axis and are opposed to each other:

Fh_{net}=F-F_r=50 N - 15N=35N

d) The net force acting on the book is the vector sum of all forces in all axes. The normal and the weight cancel each other in the y-axis, so our resulting force is the x-axis net force, computed as above:

F_n=35N in the x-axis

e) Following Newton's second law, the acceleration is calculated as

a=\frac{F_t}{m}=\frac{35N}{4Kg}=8.75m/sec^2

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3 years ago
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Step2247 [10]

Answer:

Energy loss per minute will be 120\times 10^{26}j

Explanation:

We have given the star produces power of 2\times 10^{26}W

We know that 1 W = 1 J/sec

So 2\times 10^{26}W=2\times 10^{26}J/sec

Given time = 1 minute = 60 sec

So the energy loss per minute =2\times 10^{26}\times 60=120\times 10^{26}j

We multiply with 60 we have to calculate energy loss per minute

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