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saul85 [17]
3 years ago
15

What is the general equation for burning a source

Physics
1 answer:
Shtirlitz [24]3 years ago
3 0

Answer:

The products of a complete combustion reaction include carbon dioxide (CO2) and water vapor (H2O). The reaction typically gives off heat and light as well. The general equation for a complete combustion reaction is: Fuel + O2 → CO2 + H2O.

Explanation:

The products of a complete combustion reaction include carbon dioxide (CO2) and water vapor (H2O). The reaction typically gives off heat and light as well. The general equation for a complete combustion reaction is: Fuel + O2 → CO2 + H2O.

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An electron is traveling with initial kinetic energy K in a uniform electric field. The electron comes to rest momentarily after
geniusboy [140]

Answer:

E=\frac{K}{ed}

Explanation:

We are given that

Initial kinetic energy of an electron=K

Distance=d

Final velocity=v=0

Charge,q=-1e

We have to find the magnitude of electric field.

Work done=Force\times displacement

Using the formula

Work done=qE\times d=-eEd

Using work energy theorem

Work done=Final K.E-Initial K.E=0-K

Work done=-K

Substitute the values

-K=-eEd

K=eEd

E=\frac{K}{ed}

Hence, the magnitude of the electric field=E=\frac{K}{ed}

3 0
3 years ago
What occurs when two Stars collide into each other?
ddd [48]

Answer:

A stellar collision.

Explanation:

A stellar collision is the coming together of two stars caused by stellar dynamics within a star cluster, or by the orbital decay of a binary star due to stellar mass loss or gravitational radiation, or by other mechanisms not yet well understood.

5 0
3 years ago
Read 2 more answers
If a sample emits 2000 counts per second when the detector is 1 meter from the sample, how many counts per second would be obser
Alona [7]

Answer:

<h2><em>6000 counts per second</em></h2>

Explanation:

If a sample emits 2000 counts per second when the detector is 1 meter from the sample, then;

2000 counts per second = 1 meter ... 1

In order to know the number of counts per second that would be observed when the detector is 3 meters from the sample, we will have;

x count per second = 3 meter ... 2

Solving the two expressions simultaneously for x we will have;

2000 counts per second = 1 meter

x counts per second = 3 meter

Cross multiply to get x

2000 * 3 = 1* x

6000 = x

<em></em>

<em>This shows that 6000 counts per second would be observed when the detector is 3 meters from the sample</em>

5 0
3 years ago
When an object is at a distance of twice the focal length from a concave lens, the image produced is virtual and smaller than th
Elenna [48]
The right answer for the question that is being asked and shown above is that: "The image produced is virtual and of the same size as the object." the image if the object is shifted closer to the lens to a point one focal length away from it is that The image produced is virtual and of the same size as the object.<span>
</span>
3 0
3 years ago
Read 2 more answers
particle of mass 59 g and charge 51 µC is released from rest when it is 32 cm from a second particle of charge −14 µC. Determine
AleksAgata [21]

Answer:

The initial acceleration of the 59g particle is 1062.7\frac{m}{s^{2}}

Explanation:

Newton's second laws relates acceleration (a), net force(F) and mass (m) in the next way:

F=ma (1)

We already know the mass of the particle so we should find the electric force on it to use on (1), the magnitude of the electric force between two charged objects by Columb's law is:

F=k\frac{\mid q_{1}q_{2}\mid}{r^{2}}

with q1 and q2 the charge of the particles, r the distance between them and k the constant k=9.0\times10^{9}\,\frac{Nm^{2}}{C^{2}}. So:

F=(9.0\times10^{9})\frac{\mid (51\times10^{-6})(-14\times10^{-6})\mid}{0.32^{2}}

F=62.7 N

Using that value on (1) and solving for a

a=\frac{F}{m}=\frac{62.7}{0.059}=1062.7\frac{m}{s^{2}}

4 0
3 years ago
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