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asambeis [7]
3 years ago
12

A firefighter applies a force of 2 N to lift the fire hose 3 meters up a ladder to put out a fire. How much work has the firefig

hter done?
6 N




6 J




5 N
Chemistry
2 answers:
asambeis [7]3 years ago
7 0

Answer:

The work is 6 J (Joule)

Explanation:

Work is defined as the force that is applied on a body to move it from one point to another. When a net force is applied to the body or a system and this produces displacement, then that force is said to perform mechanical work.

In the International System of Units, work is measured in Joule. Joule is equivalent to Newton per meter.

The work is equal to the product of the force by the distance (displacement) and by the cosine of the angle that exists between the direction of the force and the direction that travels the point or the object that moves.  

Work=Force*distance*cos(angle)

In this case, the angle between the force and the displacement is 0 degrees. This is because the force and displacement are in the same direction. So the cosine of 0 degrees will have a value of 1.

Also, in this case the force has a value of 2 N and the displacement or distance is 3 meters. So the work is:

Work= 2 N* 3 meters* cos(0)

Work= 2 N* 3 meters* 1

Work= 6 Joule

<u><em>The work is 6 J (Joule)</em></u>

Paraphin [41]3 years ago
5 0

Answer:

6 J

Explanation: Work done(J) = Force(N) × Distance(m) along the direction of force.

Given, Force=2 N, Distance=3 meters

∴ Wok done (J) = 2 × 3 = 6 J.

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A 250 cm^3 solution containing 1,46 g of sodium chloride is added to an excess of silver nitrate solution. The reaction is given
faust18 [17]

Answer:

The mass of the precipitate  that AgCl is 3.5803 g.

Explanation:

a) To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

We are given:

Mass of solute (NaCl) = 1.46 g

Molar mass of sulfuric acid = 58.5 g/mol

Volume of solution = 250 cm^3 =250 mL

1 cm^3= 1 ml

Putting values in above equation, we get:

\text{Molarity of solution}=\frac{1.46g\times 1000}{58.5g/mol\times 250}\\\\\text{Molarity of solution}=0.09982 M

0.09982 M is the concentration of the sodium chloride solution.

b) NaCl (aq)+AgNO_3 (aq)\rightarrow AgCI(s)+NaNO_3(aq)

Moles of NaCl = \frac{1.46 g}{58.5 g/mol}=0.02495 mol

according to reaction 1 mol of NaCl gives 1 mol of AgCl.

Then 0.02495 moles of NaCl will give:

\frac{1}{1}\times 0.02495 mol=0.02495 mol of AgCl

Mass of 0.02495 moles of AgCl:

0.02495 mol\times 143.5 g/mol=3.5803 g

The mass of the precipitate  that AgCl is 3.5803 g.

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