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ivanzaharov [21]
3 years ago
14

Balance the following chemical equations

Physics
1 answer:
IrinaVladis [17]3 years ago
4 0

Answer:

Enter an equation of a chemical reaction and click 'Balance'. The answer will appear below

Always use the upper case for the first character in the element name and the lower case for the second character.

To enter an electron into a chemical equation use {-} or e

To enter an ion specify charge after the compound in curly brackets: {+3} or {3+} or {3}. Example: Fe{3+} + I{-} =...

Substitute immutable groups in chemical compounds to avoid ambiguity. For instance equation C6H5C2H5 + O2 = C6H5OH + CO2...

Explanation:

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What is the mass of an object if it took 270 J of work to move it 15 meters? Please help!
maxonik [38]
That depends upon the direction the object moves
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3 years ago
A box of unknown mass is sliding with an initial speed vi = 5.60 m/s across a horizontal frictionless warehouse floor when it en
Maksim231197 [3]

We know that the Delta E + W(Work done by non-conservative forces) = 0 (change of energy)

In here, the non-conservative force is the friction force where f = uN (u =kinetic friction coefficient) 

W= f x d = uNd ; N=mg 
Delta E = 1/2 mV^2 -1/2mVi^2 

umgd + 1/2mV^2 - 1/2mVi^2 = 0 (cancel out the m term) 

This will then give us: 

1/2Vi^2-ugd = 1/2V^2 

V^2 = Vi^2 - 2ugd

So plugging in our values, will give us:

V= Sqrt (5.6^2 -2.3^2)

=sqrt (26.07)

= 5.11 m/s 

 

6 0
3 years ago
A private aviation helicopter's main rotor blades rotate at approximately
Arisa [49]

Answer: 7.5 rev/s

Explanation:

We are given the angular velocity \omega a helicopter's main rotor blades:

\omega=450 rpm=450 \frac{rev}{min}

However, we are asked to express this \omega in the International Systrm (SI) units. In this sense, the SI unit for time is second (s):

\omega=450 \frac{rev}{min} \frac{1 min}{60 s}

\omega=7.5 \frac{rev}{s}

4 0
3 years ago
When work is done, an object changes its
STatiana [176]

Answer:

energy is added to it

Explanation:

8 0
3 years ago
A point charge q1=+5.00nC is at the fixed position x=0, y=0, z=0. You find that you must do 8.10×10−6J of work to bring a second
Maru [420]

The value of the second charge is 1.2 nC.

<h3>Electric potential</h3>

The work done in moving the charge from infinity to the given position is calculated as follows;

W = Eq₂

E = W/q₂

<h3>Magnitude of second charge</h3>

The magnitude of the second charge is determined by applying Coulomb's law.

E = \frac{kq_2}{r^2} \\\\\frac{kq_2}{r^2} = \frac{W}{q_2} \\\\kq_2^2 = Wr^2\\\\q_2^2 = \frac{Wr^2}{k} \\\\q_2 = \sqrt{\frac{Wr^2}{k} } \\\\q_2 =  \sqrt{\frac{(8.1 \times 10^{-6}) \times (0.04)^2}{9\times 10^9} } \\\\q_2 = 1.2 \times 10^{-9} \ C\\\\q_2 = 1.2 \ nC

Thus, the  value of the second charge is 1.2 nC.

Learn more about electric potential here: brainly.com/question/14306881

7 0
2 years ago
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