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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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What is the audible range of human hearing
Phantasy [73]

Answer:

on an average, <em>2</em><em>0</em><em>H</em><em>z</em><em> </em><em>to</em><em> </em><em>2</em><em>0</em><em>k</em><em>H</em><em>z</em>

8 0
3 years ago
If 28.0 J of work is done by an external force to move a charge from a potential of 8.0 V 8.0 V to a potential of 4.0 V , 4.0 V,
jonny [76]

Answer:

Change in electric potential energy is -28.0 J

Explanation:

Electric potential energy is defined as the work is done to move a charge particle from one position to another in space in the presence of other charge particle or electric potential.

                                               OR

Electric potential energy is also equal to the change in the configuration of the charge particles.

Thus,

Change in electric potential energy = - Work Done

According to the problem, Work Done is equal to 28 J. Thus,

Change in electric potential energy = -28 J  

5 0
3 years ago
What provides the force that causes magma to erupt to the surface
olga2289 [7]
Geothermal energy provides force that causes magma to erupt to the surface
3 0
3 years ago
What is being transferred as you do work?<br> A. Energy<br> B. Power<br> C. Heat<br> D. Strength
r-ruslan [8.4K]

\huge\bold{\purple{\bold{⚡A. Energy⚡}}}

\huge\underline\mathtt\colorbox{cyan}{Reason:}

When a force causes a body to move, work is done on the object by the force. Work is the measure of the energy transfer when a force 'F' moves an object through a distance 'd'. So we say that energy is transferred from one energy store to another when work is done, and therefore, energy transferred = work done.

4 0
3 years ago
Read 2 more answers
Derive an expression for the gravitational potential energy U(r) of the object-earth system as a function of the object's distan
Drupady [299]

Answer:

U(r)=-\frac{Gm_Emr^2}{2R^3_E}

Explanation:

We are given that

Gravitational force=F_g=\frac{Gm_Emr}{R^3_E}

r=0,U(0)=0

We know that

Gravitational potential energy=-\int F_gdr

U(r)=-\int\frac{Gm_Emr}{R^3_E}dr

U(r)=-\frac{Gm_Em}{R^3_E}\times \frac{r^2}{2}+C

Substitute r=0 ,U(0)=0

0=0+C

C=0

Substitute the value

U(r)=-\frac{Gm_Emr^2}{2R^3_E}

4 0
3 years ago
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