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Radda [10]
3 years ago
14

What part of this equation represents the products? 2NH3 + 4O2?2NO + 3H2O

Physics
1 answer:
oee [108]3 years ago
6 0
C.
remember
reactants ----> products
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glass shattering in to pieces bubblegum being stretched a piece of metal expanding due to heat These are all examples of Questio
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I think it’s physical change, because the objects aren’t changing into another substance, just changing their physical shape. Hope this helps!
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Please really need help with this one <br> A <br> B<br> C<br> D
Stella [2.4K]

d thought i wouldnt trust me on it

7 0
3 years ago
A car starts from rest at the top of a hill with 45 J of gravitational
kherson [118]

Answer:

<em>The car will be moving at 5.48 m/s at the bottom of the hill</em>

Explanation:

<u>Principle of Conservation of Mechanical Energy</u>

In the absence of friction, the total mechanical energy is conserved. That means that

E_m=U+K is constant, being U the potential energy and K the kinetic energy

U=mgh

\displaystyle K=\frac{mv^2}{2}

When the car is at the top of the hill, its speed is 0, but its height h should be enough to produce the needed speed v down the hill.

The Kinetic energy is then, zero. When the car gets enough speed we assume it is achieved at ground level, so the potential energy runs out to zero but the Kinetic is at max. So the initial potential energy is transformed into kinetic energy.

We are given the initial potential energy U=45 J. It all is transformed to kinetic energy at the bottom of the hill, thus:

\displaystyle \frac{mv^2}{2}=45

Multiplying by 2:

\displaystyle mv^2=90

Dividing by m:

\displaystyle v^2=\frac{90}{m}

Taking square roots:

\displaystyle v=\sqrt{\frac{90}{m}}

\displaystyle v=\sqrt{\frac{90}{3}}

v=\sqrt{30}

v = 5.48 m/s

The car will be moving at 5.48 m/s at the bottom of the hill

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3 years ago
Would the electron released from rest move to a region of higher electrical potential or lower electrical potential
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Answer:

he electron moves in the direction of a greater electric potential

Explanation:

Electric potential and electric field are related

          V = -∫ E .ds

t

he electric force is related to the electric field

           F = q E

           F = q (-dV / ds)

since the electron has charge a = 1.6 1019 C

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therefore the electron moves in the direction of a greater electric potential

8 0
3 years ago
What happens when an atom is exposed to an energy source such as electricity?
Alex73 [517]
A) The electrons may jump to a higher energy level.
4 0
3 years ago
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