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4 years ago

The efficiency of a machine can be increased by decreasing distance. decreasing force. reducing friction. reducing work.

Chemistry

2 answers:

crimeas [40]4 years ago

5 0

Answer:

Reducing friction

Explanation:

The efficiency of a machine is a how the input energy is converted in output work

Then in the middle you have things consuming the input energy and not going to the output work, the best you can do is reducre those things consuming, for a machine it can be friction, heat, vibration or loose parts.

For that reason Reducing friction is the answer

xeze [42]4 years ago

5 0

Answer: Reducing friction

Explanation:

Friction refers to the resistance that a surface encounters while moving on the surface of another. Machines often posses moving parts such as rollers. Those parts move on other surfaces generating friction. Friction in machine parts leads to useless work and reduces the amount of input work used by the machine to do useful work. The efficiency of a machine has to do with its ability to convert input work into output work. This efficiency is greatly reduced by friction in the moving parts of the machine. Hence friction decreases the efficiency (work done by a machine)

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2 years ago

A 3.00-L flask is filled with gaseous ammonia, NH3. The gas pressure measured at 27.0 ∘C is 2.55 atm . Assuming ideal gas behavi

Whitepunk [10]

Answer : The mass of ammonia present in the flask in three significant figures are, 5.28 grams.

Solution :

Using ideal gas equation,

$PV=nRT\\\\PV=\frac{w}{M}\times RT$

where,

n = number of moles of gas

w = mass of ammonia gas = ?

P = pressure of the ammonia gas = 2.55 atm

T = temperature of the ammonia gas = $27^oC=273+27=300K$

M = molar mass of ammonia gas = 17 g/mole

R = gas constant = 0.0821 L.atm/mole.K

V = volume of ammonia gas = 3.00 L

Now put all the given values in the above equation, we get the mass of ammonia gas.

$(2.55atm)\times (3.00L)=\frac{w}{17g/mole}\times (0.0821L.atm/mole.K)\times (300K)$

$w=5.28g$

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