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lesantik [10]
3 years ago
9

If measuring experimental results, what can you predict about the work output of a 1200 watt hair dryer? the work output of the

hairdryer will be less than the work input the work output of the hairdryer will be equal to the work input the work output of the hairdryer will be greater than the work input the work output of the hairdryer cannot be predicted in the real world
Chemistry
2 answers:
eimsori [14]3 years ago
6 0

Answer: first option, the work output of the hairdryer will be less than the work input.


Explanation:


1) The work output measured in watts is the power of hair dryer measured in joules per second.


2) The hair dryer converts electrical energy from the wall outlet to mechanical and thermal energy: hot wind.


3) Nevertheless, you can never expect a 100% efficiency of the machines: due to friction, some energy is converted into useless energy.


So, efiiviency = power output / power input< 1 ⇒


power output = work output / time


input power = work input / time


⇒ work output / work input < 1


⇒ work output < work input.


Which is the first option: the work output of the hairdryer will be less than the work input

topjm [15]3 years ago
3 0
<span>The work output of a 1200 wwtt hair dryer will be less than the work input. Work input is equal to Force times distance. To solve the mechanical efficiency it will be work output divided by work input times 100%. It is a simple machine so the work output is less than the work input which means the work. In addition, work output is overcomes friction causing it to lose its energy but there is still work done . Work input is equal to the work output plus the work done to overcome the friction.</span>
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masya89 [10]

Answer:

This involves negatively charged particles (electrons) jumping to positively charged objects. When you rub the balloons against the fabric they become negatively charged. They take some of the electrons from the fabric and leave them positively charged.

Explanation:

Negative charges attract to positive charges. If a balloon is not rubbed with the wool cloth, it has an equal amount of negative to positive charges, so it will attract to a rubbed balloon. When both balloons are rubbed with the wool cloth, the both receive negative charges, so they will repel each other.

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3 years ago
A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 ?C. The initial concentrations of Pb2+ and Cu2+ are
VARVARA [1.3K]

Answer:

a) Ecell = 0.5123 V

b) Ecell =  0.4695 V

c) [Pb2 +] = 4.75 M

Explanation:

a)

The reaction at the cathode is represented as follows:

Cu2 + + 2e- -> Cu (s) Eocathode = 0.34 V

The reaction at the anode is equal to:

Pb (s) -> Pb2 + + 2e- Eoanode = -0.13 V

The number of moles of the electrons that are involved is equal to n = 2

Standard cell potential equals Eo = Eocathode - Eoanode = 0.34 V- (-0.13 V) = 0.47 V

 The initial cell potential can be calculated with the following formula:

Ecell = Eocell - - 0.0592 / n) log ([(Pb2 +)] / [(Cu2 +)]) = 0.47 - (0.0592 / 2) log (0.052 / 1.4) = 0.5123 V

b)

The reaction in the cell is equal to:

Cu2 + + Pb (s) -> Cu (s) + Pb2 +

The concentration of Cu2 that gives the exercise is equal 0.2 M

Therefore, the change in concentration for Cu2 + is equal to:

Cu2 + = 1.4 M - 0.2 M = 1.2 M

We use the formula from part a)

Ecell = Eocell - (0.0592 / n) log ([(Pb2 +)] / [(Cu2 +)]) = 0.47 - (0.0592 / 2) log (1,252 / 1.2) = 0.4695 V

c)

To find the concentration of Pb2 + when there is a potential change in the cell of 0.37 V, we must clear the concentration of Pb2 + from the following formula:

Eccell = Echocell - (0.0592 / n) log (([Pb2 +]) / ([Cu2 +]))

0.0296 log ([Pb2 +] / [Cu2 +]) = (Eocélula - Ecélula / 0.0296)

Clearing Pb2 +:

[Pb2 +] = 4.75 M

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The concentration of species in 500 mL of a 2.104 M solution of sodium sulfate is ________ M sodium ion and ________ M sulfate i
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Answer:

The concentration of species in 500 mL of a 2.104 M solution of sodium sulfate is 4.208 M sodium ion and 2.104 M sulfate ion.  (option E)

Explanation:

Step 1: Data given

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The concentration of species in 500 mL of a 2.104 M solution of sodium sulfate is 4.208 M sodium ion and 2.104 M sulfate ion.  (option E)

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