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3 years ago

Can you make the work output of a machine greater than the work input?

Physics

2 answers:

ExtremeBDS [4]3 years ago

5 0

Answer:

It is<em> impossible</em> to construct a machine which produces the <em>work output greater than the work input.</em>

Let us consider the II law of thermodynamics.

According to Kelvin Plank's statement any engine/machine does not give hundred percent efficiency. And violating the PMM-II(Perpetual motion of machine II kind), Always some amount of energy transferred to the sink or surroundings.

Therefore

W(ouput) = Q₁-Q₂

There are many reasons to lower the work output, just for an example friction between the mating parts reduces the work output.

VLD [36.1K]3 years ago

5 0

Answer:

Explanation:

A machine makes are work easier. The mechanical advantage of a machine tells us the ability of machine to multiply the input force. No machine is 100% efficient. It means it cannot convert all the input completely to output, let alone give a greater output. Some of the work that is input into the machine is always lost in overcoming friction produced.

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finlep [7]

Answer:

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Explanation:

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2 years ago

Suppose you increase your walking speed from 6 m/s to 14 m/s in a period of 2 s. What is your acceleration? m/s2

saveliy_v [14]

4.5m/s you count the number between 6 and 14 and get 9 do 9 divided by 2 and you get 4.5 m/s increase. hope this helps.

8 0

3 years ago

Charge is placed on two conducting spheres that are very far apart and connected by a long thin wire. The radius of the smaller

mrs_skeptik [129]

Answer:

surface charge density on each sphere is $440 \times 10^{-9}$ C

Explanation:

given data

radius of smaller sphere = 5 cm

radius of larger sphere is 12 cm

electric field at surface of larger sphere = 660 kV/m = 660 × 1000 v/m

solution

we apply here electric field formula that is express as

E = $(\frac{1}{4\pi\epsilon })\times (\frac{Q_{1} }{R^{2} } )$ .................1

put here value

660000 = $9 \times 10^9 \times \frac{Q1}{0.12^2}$

Q1 = 1056 × $10^{-9}$

and

here field inside a conductor is zero so that electric potential ( V ) is constant

$\frac{Q{1} }{R} = \frac{Q{2} }{r}$ ..................2

so Q2 will be

Q2 = $\frac{5}{12} \times 1056 \times 10^{-9}$

Q2 = $440 \times 10^{-9}$ C

6 0

3 years ago

A 20" round duct is 275' in length. The duct is carrying 2,900 CFM at a friction loss of 0.12 inches WG per 100 feet.

ivolga24 [154]

50500000 FACTION : uo/80

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2 years ago

Tres litros de oxigeno gaseoso a 15 grados centígrados y a presión atmosférica (1atm), se lleva a una presión de 10mm de Hg. ¿ c

lana [24]

Answer:

Three liters of gaseous oxygen at 15 degrees Celsius and at atmospheric pressure (1atm), is brought to a pressure of 10mm Hg. What will be the volume of the gas now if the temperature has not changed?

Explanation:

Given that, the temperature is constant

Then, using gay Lussac law

P1•V1 / T1 = P2•V2 / T2

Since temperature is constant

Then, T1 = T2 = T and they cancels out

So, we are left with

P1•V1 = P2•V2

Given that, .

Initial volume

V1 = 3 litres.

Initial pressure

P1 = 1atm = 101325 Pa

Final pressure

P2 = 10mmHg = 1333.22 Pa

Then, we want to find the final volume V2

Make V2 subject of formula.

V2 = V1•P1 / P2

V2 = 3 × 101325 / 1333.22

V2 = 288 litres

So, the final volume is 288 litres.

In Spanish

Dado que la temperatura es constante

Luego, usando la ley gay de Lussac

P1 • V1 / T1 = P2 • V2 / T2

Como la temperatura es constante

Entonces, T1 = T2 = T y se cancelan

Entonces, nos quedamos con

P1 • V1 = P2 • V2

Dado que, .

Volumen inicial

V1 = 3 litros.

Presión inicial

P1 = 1atm = 101325 Pa

Presión final

P2 = 10 mmHg = 1333.22 Pa

Entonces, queremos encontrar el volumen final V2

Hacer V2 sujeto de fórmula.

V2 = V1 • P1 / P2

V2 = 3 × 101325 / 1333.22

V2 = 288 litros

Entonces, el volumen final es de 288 litros

8 0

3 years ago