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rjkz [21]
3 years ago
7

Describe what the effect of increasing the power of a camera would have on the battery life

Physics
1 answer:
patriot [66]3 years ago
8 0

Answer:

. Cut Down on the LCD

The biggest battery drain in a camera is the LCD – both the rear screen and the electronic viewfinder. This is the big reason why DSLRs almost always have longer battery life specifications than mirrorless cameras – the optical viewfinder lets you skip LCDs altogether.

However, if you use your DSLR in live view, you’ll notice that its battery life slides dramatically. Side by side against a mirrorless camera, there’s actually a good chance it will die first. LCDs just take a lot of power to run.

What does this imply? Quite simply, you should always do what you can to cut down on LCD usage when your battery is running low.

For DSLR users, that means switching to the optical viewfinder. For mirrorless photographers, it means turning off the camera frequently, or setting it so the viewfinder only activates when you hold it to your eye.

And regardless of the camera you use, drastically cut down on the amount of time you spend reviewing photos. Chimping has its place, but not while your battery warning is blinking red.Optimize Your Battery Saver Settings

Most cameras have menu options designed to improve battery life and maximize your shooting time. For example, the “metering timeout” setting lets you select how long you want the camera to wait during inactivity before shutting off its metering system.

Beyond that, a number of cameras today have an “Eco mode” that minimizes power consumption from the camera’s LCD. On the Canon EOS R, for example, Eco mode dims and then turns off the LCD when not in use, improving your battery life significantly – from 370 to 540 shots per charge, according to Canon’s official specifications.

It’s also important to note that mirrorless cameras are generally more efficient using the rear LCD than the electronic viewfinder. In terms of the EOS R again, Canon only rates 350 shots using the EVF, with no Eco mode to improve it. On the Sony side of things, the new A7R IV is rated for 530 shots via the viewfinder and 670 via the rear LCD.

If none of that applies to you, one option at your disposal is always to lower the brightness of your rear LCD. It might make photography a bit trickier in bright conditions, but the payoff is getting the shot rather than missing it completely due to a dead battery.

Other camera settings and extras that harm battery life include:

Image stabilization (both in-body and in-lens)

Popup flash

Bluetooth and WiFi

Most external accessories: GPS dongles, lightning triggers, wireless remote releases, shotgun mics, etc.

Sometimes, these capabilities are essential for your photo, so it’s worth the battery life sacrifice. But if you’re down to your last bar, double check to ensure that you’re not using any of the above settings or accessories without good reason.

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Ruby, a variety of the mineral corundum is in the trigonal crystal system, with hexagonal scalenohedra crystals
7 0
3 years ago
A baseball is batted from a height of 1.09 m with a speed of
kobusy [5.1K]

(a) The horizontal and vertical components of the ball’s initial velocity is 37.8 m/s and 12.14 m/s respectively.

(b) The maximum height above the ground reached by the ball is 8.6 m.

(c) The distance off course the ball would be carried is 0.38 m.

(d) The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

<h3>Horizontal and vertical components of the ball's velocity</h3>

Vx = Vcosθ

Vx = 39.7 x cos(17.8)

Vx = 37.8 m/s

Vy = Vsin(θ)

Vy = 39.7 x sin(17.8)

Vy = 12.14 m/s

<h3>Maximum height reached by the ball</h3>

H = \frac{v^2 sin^2(\theta)}{2g} \\\\H = \frac{(39.7)^2 \times (sin17.8)^2}{2(9.8)} \\\\H = 7.51 \ m

Maximum height above ground = 7.51 + 1.09 = 8.6 m

<h3>Distance off course after 2 second </h3>

Upward speed of the ball after 2 seconds, V = V₀y - gt

Vy = 12.14 - (2x 9.8)

Vy = - 7.46 m/s

Horizontal velocity will be constant = 37.8 m/s

Resultant speed of the ball after 2 seconds = √(Vy² + Vx²)

V = \sqrt{(-7.46)^2 + (37.8)^2} \\\\V = 38.53 \ m/s

<h3>Resultant speed of the ball and crosswind</h3>

V = \sqrt{38.52^2 + 4^2} \\\\V = 38.72 \ m/s

<h3>Distance off course the ball would be carried</h3>

d = Δvt = (38.72 - 38.53) x 2

d = 0.38 m

The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

Learn more about projectiles here: brainly.com/question/11049671

5 0
1 year ago
If a weight hanging on a string of length 5 feet swings through 5^\circ on either side of the vertical, how long is the arc thro
Zielflug [23.3K]
<span>First we can find the circumference of the whole circle with a radius of 5 feet. circumference = 2 pi radius circumference = (2 pi) (5 feet) circumference = (10 pi) feet From one high point to the other high point, the string moves through an angle of 10 degrees. Since a full circle is 360 degrees, this angle is 1/36 of a full circle. Therefore, the arc length is 1/36 of the whole circumference. arc length = (1/36) (circumference) arc length = (1/36) (10 pi) feet arc length = 0.873 feet</span>
7 0
3 years ago
Which image shows both potential and kinetic energy?
Natasha2012 [34]

Answer:

The one in the middle

Explanation: i listened to the other person and i got it wrong, this is the answer for edge2020 sience review on energy!!!!

trust me its the middle one!!!!!

And everyone if ur not sure, like 100% sure about an answer dont answer at all cuz for 1: ur taking up a spot for others to answer. for 2: you could make people wrong. And for 3: its annoying. And 4: it makes stuff like this happen!

<u>NOT ARGUEING IM JUST PUTTING MY THOUGHTS AND OPINIONS OUT THERE ;)</u><em> many thanks.</em>

4 0
3 years ago
Read 2 more answers
A closely wound rectangular coil of 80 turns has dimensions 25.0 \rm cm by 40.0 \rm cm. The plane of the coil is rotated from a
Pepsi [2]

Answer:

88.3

Explanation:

Emf in a rotating coil is given by rate of change of flux:

E= dФ/dt=(NABcos∅)/ dt

N: number of turns in the coil= 80

A: area of the coil= 0.25×0.40= 0.1

B: magnetic field strength= 1.1

Ф: angle of rotation= 90- 37= 53

dt= 0.06s

E= (80 × 0.4× 0.25×1.10 × cos53)/0.06= 88.3V

4 0
3 years ago
Read 2 more answers
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