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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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PLEASE HELP ASAP! WILL GIVE BRANLIEST! Pay attention to what happens to the things around you as you go about your day. Describe

umka2103 [35]

Answer:

jumping, pulling a elastic band, bouncing a ball

Explanation:

when you jump, your legs apply a force to the ground, and the ground applies and equal and opposite reaction force that propels you into the air.

When we pull an elastic band, it automatically returns to its original position. The more you pull the more force it generates. This is the same when you pull or compress a spring. The action (applied force) is stored as energy and is released as a reaction with an equal and opposite force

A ball is able to bounce because of the reaction from the ground. If there was no reaction then the ball would not bounce but rather stick to the ground.

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3 years ago

A 50 g ice cube can slide without friction up and down a 30â slope. the ice cube is pressed against a spring at the bottom of th

vfiekz [6]

Fk= mgcostheta * coefficient of friction
Fk= (.05kg)(9.80)(cos30)(.20)=.085 N

change in thermal energy= Fk* distance covered up slope (d)
change in thermal energy = .085 * d

the final height so y=dsin30

Kf + Uf + change in thermal energy = Ki + Ui

(.05kg)(9.80)(dsin30) + (.085d) = (.5)(25 N/m)(.1m)^2

d= .379 m = 37.9 cm

4 0

3 years ago

A 200 g piece of iron is heated to 100°C. It is then dropped into water to bring its temperature down to 22°C. What is the amoun

Amiraneli [1.4K]

The piece of iron is brought from a temperature of 100C to a temperature of 22C. In this process, the heat released by this piece of iron is
$Q=m C_s \Delta T$
where $m=200 g$ is the mass of the piece of iron, $C_s = 0.444 J/(g C)$ is the iron's specific heat, and the temperature variation is $\Delta T = 22^{\circ}-100^{\circ}=-78^{\circ}C$ . Using these values, the amount of heat released by the iron is
$Q=(200 g)(0.441 J/(gC))(-78C)=-6926J=-6.9 kJ$
With negative sign because it is amount of heat released. This heat is then trasferred to the water, so the amount of heat absorbed by the water is Q=6.9 kJ.

8 0

3 years ago

Read 2 more answers

Cyclist A is moving at 20.0 m/s whereas cyclist B is moving at 12.0 m/s in the same direction and is initially ahead ofA. When t

stealth61 [152]

Answer:

$v_{fA} = 28 \frac{m}{s}$

Explanation:

The kinematic parameters from the moment the two cyclists begin to accelerate until they meet are:

Initial parameters:

$v_{oA} = 20 m/s$ : Initial speed of cyclist A

$v_{oB} = 12 m/s$ : Initial speed of cyclist B

Final parameters:

$v_{fB} = 20 m/s$ : Final speed of cyclist B

$d_{A} = d_{B}$

distance of cyclist A = distance of cyclist B

$t_{A} =t_{B} = 12s$

time of cyclist A = time of cyclist B

Cyclist B Kinematics

$v_{fB} = v_{oB} +a_{B} *t\\$

$36= 12 + a_{B} *12$

$a_{B} = \frac{36-12}{12}$

$a_{B} = 2 \frac{m}{s^{2} }$

$d_{B} =( v_{oB})*( t)+( \frac{1}{2} )*(a_{B})* (t)^{2}$

$d_{B} =( 12*(12)+( \frac{1}{2} )*(2)* (144)}$

$d_{B} = 288m$

Cyclist A Kinematics

$d_{A} =( v_{oA})*( t)+( \frac{1}{2} )*(a_{A})* (t)^{2}$

$d_{A} =(20*( 12)+( \frac{1}{2} )*(a_{A})* 144}$

$d_{A} = 240 + 72*(a_{A})$

$288=240+72*(a_{A} )$

$a_{A} = \frac{288-240}{72}$

$a_{A} = 0.67 \frac{m}{s^{2} }$

$v_{fA} = v_{oA} + a_{A} * t$

$v_{fA} = 20 + 0.67*12$

$v_{fA} = 28 \frac{m}{s}$

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3 years ago

Can things of aluminum have a greater mass than things made of iron?

nikklg [1K]

Yes, if the object is larger. while a solid block of iron will have a greater mass than a block of aluminum of the same size, a much larger block of aluminum can have greater mass than the smaller block of iron.

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3 years ago