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Temka [501]
3 years ago
5

Please only answer if you are positive. Thanks! Brainliest if correct!

Physics
2 answers:
kifflom [539]3 years ago
6 0
The answer is mountain peak
baherus [9]3 years ago
5 0

Answer:

They create a Mountain Peak

Explanation:

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kotykmax [81]
Accuracy?

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3 0
3 years ago
How much potintial energy is in a closed system where the height of the object is 2 m, and the mass is 10 kg?
Lorico [155]

Answer:

D. 196 J

Explanation:

PE=mgh=10×9.8×2=196J

6 0
2 years ago
Read 2 more answers
HELLLPPP FAST PLS
r-ruslan [8.4K]

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Oh I had this question before! :)

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Anything with a sour taste is acidic

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3 0
3 years ago
What is the definition of acceleration
PtichkaEL [24]
Acceleration is the way the motion is changing.
4 0
3 years ago
Please help me
qaws [65]

-- We know that the y-component of acceleration is the derivative of the
y-component of velocity.

-- We know that the y-component of velocity is the derivative of the
y-component of position.

-- We're given the y-component of position as a function of time.

So, finding the velocity and acceleration is simply a matter of differentiating
the position function ... twice.

Now, the position function may look big and ugly in the picture.  But with the
exception of  't' , everything else in the formula is constants, so we don't even
need any fancy processes of differentiation.  The toughest part of this is going
to be trying to write it out, given the text-formatting capabilities of the wonderful
envelope-pushing website we're working on here.

From the picture . . . . . y (t) = (1/2) (a₀ - g) t² - (a₀ / 30t₀⁴ ) t⁶

First derivative . . . y' (t) = (a₀ - g) t  -  6 (a₀ / 30t₀⁴ ) t⁵  =  (a₀ - g) t  -  (a₀ / 5t₀⁴ ) t⁵

There's your velocity . . . /\ .

Second derivative . . . y'' (t) = (a₀ - g) -  5 (a₀ / 5t₀⁴ ) t⁴ = (a₀ - g) -  (a₀ /t₀⁴ ) t⁴

and there's your acceleration . . . /\ .
That's the one you're supposed to graph.

a₀ is the acceleration due to the model rocket engine thrust
     combined with the mass of the model rocket
'g' is the acceleration of gravity ... 9.8 m/s² or 32.2 ft/sec²
t₀  is how long the model rocket engine burns

Pick, or look up, some reasonable figures for a₀ and t₀
and you're in business.

The big name in model rocketry is Estes.  Their website will give you
all the real numbers for thrust and burn-time of their engines, if you
want to follow it that far.


6 0
3 years ago
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