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Harman [31]
2 years ago
15

As a result of recent weather activity, there’s less water available for human consumption. Which biome is affected the most by

this change?
a)tundra
b)taiga
c)saltwater
d)freshwater
Physics
2 answers:
andreev551 [17]2 years ago
7 0
The correct answer for this question is this one: "d)freshwater"
<span>As a result of recent weather activity, there’s less water available for human consumption. The biome that is affected the most by this change is that <u>freshwater</u></span>
Hope this helps answer your question and have a nice day ahead.
Molodets [167]2 years ago
5 0

Answer: d)freshwater

The freshwater biome will be most likely to be affected by the weather conditions as the freshwater reservoirs accumulates water from sources like rain, flood and melting of snow. The reservoirs form are groundwater, river, lakes, ponds and others. In the absence of rain the water will not be accumulated in the biome and will effect the life forms living there.

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Describe what the effect of increasing the power of a camera would have on the battery life
patriot [66]

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.

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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.

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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:

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8 0
2 years ago
A shopping cart given an initial velocity of 2.0 m/s undergoes a constant acceleration to a velocity of 13 m/s. What is the magn
olga55 [171]

Answer:

The acceleration is a = 2.75 [m/s^2]

Explanation:

In order to solve this problem we must use kinematics equations.

v_{f} = v_{i} + a*t\\

where:

Vf = final velocity = 13 [m/s]

Vi = initial velocity = 2 [m/s]

a = acceleration [m/s^2]

t = time = 4 [s]

Now replacing:

13 = 2 + (4*a)

(13 - 2) = 4*a

a = 2.75 [m/s^2]

5 0
3 years ago
A 3600 kg rocket traveling at 2900 m/s is moving freely through space on a journey to the moon. The ground controllers find that
Nana76 [90]

Answer:

m=417.24 kg

Explanation:  

Given Data

Initial mass of rocket  M = 3600 Kg

Initial velocity of rocket vi = 2900 m/s  

velocity of gas vg = 4300  m/s

Θ = 11° angle in degrees

To find

m = mass of gas  

Solution

Let m = mass of gas    

first to find Initial speed with angle given

So

Vi=vi×tanΘ...............tan angle

Vi= 2900m/s × tan (11°)

Vi=563.7 m/s

Now to find mass

m = (M ×vi ×tanΘ)/( vg + vi tanΘ)

put the values as we have already solve vi ×tanΘ

m = (3600 kg ×563.7m/s)/(4300 m/s + 563.7 m/s)

m=417.24 kg

7 0
3 years ago
A gorilla weighs 8.00x10^2N and swings from vine to vine. As the gorilla grads a new vine, both vines make an angle of 30 degree
Yakvenalex [24]

Answer:

461.88 N  

Explanation:

Let tension in each vine be T .

The vertical component of this tension of both the vine will add up and balance the weight .

2T cos 30 = mg

2 T x cos 30 = 800

T  = 800 / (2 x cos30 )

= 461.88 N  

6 0
2 years ago
Dois carros, a e b, móveis se em uma estrada retilínea com volocidade constante,vª =20m/s e v=18 m/s , respectivamente. O carro
aalyn [17]

Answer:

t = 250 s = 4.167 s

Explanation:

Carro A

x = x₀+v*t

⇒  x = 0 + 20*t  

⇒  x = 20*t   <em>(i)</em>

Carro B

x = x₀+v*t

⇒  x = 500 + 18*t    <em>(ii)</em>

Se as equações são as mesmas, verifica-se  <em>(i) = (ii)</em>

20*t = 500 + 18*t

⇒  20*t - 18*t = 500

⇒  2*t = 500

⇒  t = 250 s = 4.167 s

6 0
3 years ago
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