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VARVARA [1.3K]
3 years ago
10

Where does energy go when an object hits the ground?

Physics
1 answer:
PolarNik [594]3 years ago
8 0
It turns into kinetic energy; as an object falls it loses potential energy and it can be transfer into friction into heat. 
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Can someone help me answer please
Andru [333]

Answer:

4=Conduction by convection by radiation.

Explanation:

Hope it will help you! It may be short but I don't know how to write it in blank aafai milayera lekha Hai blanks ma

5 0
2 years ago
As shown in the diagram below, a 1 kg rock tied to a rope is
Tatiana [17]

Answer:

Explanation:

Angular momentum has a formula of L = mvr. Fillingin:

L = (1.0)(5.0)(1.0)

L = 5.0 kg*m/s

4 0
3 years ago
a 72kg person is standing on a bathroom scale (calibrated in newtons). what is the reading of the scale when the acceleration is
Likurg_2 [28]

Answer:

F=ma

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Explanation:

i think that's what your doing but I'm not sure

4 0
2 years ago
A jet of water of cross section area and velocity v impinges normally on a stationary flat plate the mass per unit volume of wat
Sergio039 [100]
<h2>F = kAρv²</h2>

Explained in the attachment !

<h3>Hope it helps you!!</h3>

4 0
2 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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