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denpristay [2]
2 years ago
13

A combustion reaction is represented by __________________.

Physics
1 answer:
Ugo [173]2 years ago
8 0

Answer:

The general equation for a complete combustion reaction is: Fuel + O2 → CO2 + H2O. The burning of charcoal is a combustion reaction

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so whats the  questain?!

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newton's second law states that when a net force acts on an object, it accelerates it.Explain how it would be possible for two o
Alika [10]

Answer:

See Explanation

Explanation:

According to Newton's second law of motion, the acceleration of a body is proportional to the net external force that acts on the body.

A body accelerated when it is acted upon by an unbalanced net external force.

When the external forces acting on a body are balanced, the effect of each force is cancelled by the other hence the body is not accelerated according to Newton's second law.

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What are the major weather fronts? How do they differ from each other?
dmitriy555 [2]

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they differ each other bc it is a lot of with earth quakes and everything.

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3 years ago
Help me, i don't know the answer ​
8_murik_8 [283]

Explanation:

1)

A) Bb BB

B) 50%

2)

A) 50%

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4 0
2 years ago
An electron is placed on a line connecting two fixed point charges of equal charge but the opposite sign. The distance between t
viktelen [127]

Answer:

a)    F_net = 6.48 10⁻¹⁸ ( \frac{1}{x^2} + \frac{1}{(0.300-x)^2} ),   b) x = 0.15 m

Explanation:

a) In this problem we use that the electric force is a vector, that charges of different signs attract and charges of the same sign repel.

The electric force is given by Coulomb's law

         F =k \frac{q_2q_2}{r^2}

         

Since when we have the two negative charges they repel each other and when we fear one negative and the other positive attract each other, the forces point towards the same side, which is why they must be added.

          F_net= ∑ F = F₁ + F₂

let's locate a reference system in the load that is on the left side, the distances are

left side - electron       r₁ = x

right side -electron     r₂ = d-x

let's call the charge of the electron (q) and the fixed charge that has equal magnitude Q

we substitute

          F_net = k q Q  ( \frac{1}{r_1^2}+ \frac{1}{r_2^2})

          F _net = kqQ  ( \frac{1}{x^2} + \frac{1}{(d-x)^2} )

         

let's substitute the values

          F_net = 9 10⁹  1.6 10⁻¹⁹ 4.50 10⁻⁹ ( \frac{1}{x^2} + \frac{1}{(0.30-x)^2} )

          F_net = 6.48 10⁻¹⁸ ( \frac{1}{x^2} + \frac{1}{(0.300-x)^2} )

now we can substitute the value of x from 0.05 m to 0.25 m, the easiest way to do this is in a spreadsheet, in the table the values ​​of the distance (x) and the net force are given

x (m)        F (N)

0.05        27.0 10-16

0.10          8.10 10-16

0.15          5.76 10-16

0.20         8.10 10-16

0.25        27.0 10-16

b) in the adjoint we can see a graph of the force against the distance, it can be seen that it has the shape of a parabola with a minimum close to x = 0.15 m

4 0
3 years ago
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