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Bond [772]
3 years ago
5

What causes the lens of the camera to extend or retract

Physics
1 answer:
kotykmax [81]3 years ago
8 0
The glass has a special tint to focus as you turn it a certain way!
Brainliest pls! thx! :)

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A 0.144-kg baseball is moving toward home plate with a speed of 43 m/s when
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I would say 648858. bc yes
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3 years ago
An airplane starts from rest and accelerates at 10.8 m/s2 . what is its speed at the end of a 400 m long runway?
Cloud [144]

The final speed of an airplane is v = 92.95 m/s

The rate of change of position of an object in any direction is known as speed i.e. in other word, Speed is measured as the ratio of distance to the time in which the distance was covered.

Solution-

Here given,

Acceleration a= 10.8 m/s2 .

Displacement (s)= 400m

Then to find final speed of airplane v=?

Therefore from equation of motion can be written as,

v²=u²+ 2as

where, u is initial speed, v is final speed ,a is acceleration and s is displacement of the airplane. Therefore by putting the value of a & s in above equation and (u =0) i.e. the initial speed of airplane is zero.

v²= 2×10.8 m/s²×400m

v²=8640m/s

v=92.95m/s

hence the final speed of airplane v =92.95m/s

To know more about speed

brainly.com/question/13489483

#SPJ4

5 0
2 years ago
What are the properties of bungee gum,hmmm?
lys-0071 [83]
It’s both a solid and a liquid. It can thicken and soften depending on how it’s handled. It can be used to cover wounds to stop bleed, and used to drown enemies. Bungee Gum has the properties of both rubber and gum.
8 0
3 years ago
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Which of these is a chemical property of a substance?
My name is Ann [436]
Reactivity is a chemical property of substance
5 0
4 years ago
Col. John Stapp led the U.S. Air Force Aero Medical Laboratory's research into the effects of higher accelerations. On Stapp's f
Lerok [7]

Answer:

A.a=203.14\ \frac{m}{s^2}

B.s=397.6 m

Explanation:

Given that

speed  u= 284.4 m/s

time t = 1.4 s

here he want to reduce the velocity from 284.4 m/s to 0 m/s.

So the final speed v= 0 m/s

We know that

v= u + at

So now by putting the values

0 = 284.4 -a x 1.4     (here we take negative sign because this is the case of de acceleration)

a=203.14\ \frac{m}{s^2}

So the acceleration  while stopping will be a=203.14\ \frac{m}{s^2}.

Lets take distance travel before come top rest is s

We know that

v^2=u^2-2as

0=284.4^2-2\times 203.14\times s

s=397.6 m

So the distance travel while stopping is 397.6 m.

8 0
4 years ago
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