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Naddik [55]
3 years ago
6

Stacy travels 5 times as fast as Eric. Traveling in opposite directions, they are 336 miles apart after 4 hours. Find their rate

s of travel.
Physics
1 answer:
Vilka [71]3 years ago
6 0

Answer:

Eric travelled at 21miles per hour

while stacy travelled at 105miles per hour.

Explanation:

speed = distance/time

then distance = speedxtime.

speed = x

time = 4hrs

Eric's distance after 4hrs = 4x

since stacy is 5 times faster than eric =

stacy's distance = (5x)4 = 20x

Stacy's distance - Eric's distance = 336m

20x - 4x = 336Miles

16x = 336miles

x = 336/16

x = 21miles per hour

so Eric was travelling at 21miles per hour

While Stacy was travelling at 105 miles per hour.

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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.

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.

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8 0
3 years ago
A flea jumps straight up to a maximum height of 0.550 m . what is its initial velocity v0 as it leaves the ground?
Alexxx [7]
Since my givens are x = .550m [Vsub0] = unknown
 [Asubx] = =9.80
 
 [Vsubx]^2 = [Vsub0x]^2 + 2[Asubx] * (X-[Xsub0]

[Vsubx]^2 = [Vsub0x]^2 + 2[Asubx] * (X-[Xsub0]) 

Vsubx is the final velocity, which at the max height is 0, and Xsub0 is just 0 as that's where it starts so I just plug the rest in

0^2 = [Vsub0x]^2 + 2[-9.80]*(.550)

0 = [Vsub0x]^2 -10.78

10.78 = [Vsub0x]^2

Sqrt(10.78) = 3.28 m/s 


3 0
3 years ago
What does the principal quantum number determine?
morpeh [17]
<span>principal quantum number (n) </span>represents the relative overall energy of each orbital

Hope this helps!
4 0
3 years ago
How much acceleration can most humans stand before they pass out?
kumpel [21]
Normal humans can withstand no more than 9 g's, and even that for only a few seconds. When undergoing an acceleration of 9 g's, your body feels nine times heavier than usual, blood rushes to the feet, and the heart can't pump hard enough to bring this heavier blood to the brain.
4 0
2 years ago
A 60 g golf ball is dropped from a level of 2 m high. It rebounds to 1.5 m. How much energy is lost? Group of answer choices 0.5
bogdanovich [222]

Answer: A 60 g golf ball is dropped from a level of 2 m high. It rebounds to 1.5 m. Energy loss will be 0.29J

Explanation: To find the correct answer, we have to know more about the Gravitational potential energy.

<h3>What is gravitational potential energy?</h3>
  • The energy possessed by a body by virtue of its position in gravitational field of earth is called gravitational potential energy.
  • The gravitational potential energy of a body at a height h with respect to the height h will be,

                                          U=mgh

  • Expression for gravitational potential energy loss will be,

                                        E=U_i-U_f

<h3>How to solve the problem?</h3>
  • The total energy before the ball dropped will be,

                 U_i=mgh_i=60*10^-3kg*9.8m/s^2*2m=1.176 J

  • The total energy after when the ball rebounds to 1.5m will be,

                 U_f=mgh_f=60*10^-3kg*9.8m/s^2*1.5m=0.882J

  • The total energy loss will be,

                E=1.176-0.882=0.294J

Thus, we can conclude that, the energy loss will be,0.294J.

Learn more about the gravitational potential energy here:

brainly.com/question/28044692

#SPJ4

3 0
2 years ago
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