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1 year ago

How can you make the work output of machine greater than the work input

Physics

1 answer:

ivolga24 [154]1 year ago

5 0

Answer:

The only way to increase the work output is to increase the work input. You can't get more work out of a machine than you put into it. The moving parts of a machine must use some of the work input to overcome friction. Because of friction, the work output of a machine is always less than the work input.

Explanation:

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if a cat is running at a constant speed of 10km/h for 5 s, what is its average speed and what is its instantaneous speed at 4 s?

iris [78.8K]

Here a cat is running at constant speed which is given as 10 km/h for 5s

So here the average speed is defined as total distance moved in total time interval

so here it is given by

$v_{avg} = \frac{distance}{time}$

since

here speed of cat is constant so it will remain the same

And hence the average speed and instantaneous speed at any instant for this duration will remain the same

so here answer would be

<em>average speed = 10 km/h</em>

<em>instantaneous speed = 10 km/h</em>

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2 years ago

Uses of hydrometer in points

Arisa [49]

Answer:

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2 years ago

An insulated Thermos contains 140 cm3 of hot coffee at 85.0°C. You put in a 15.0 g ice cube at its melting point to cool the cof

Pavel [41]

Answer:

$T = 69^o C$

Explanation:

Here at thermal equilibrium we can say that thermal energy given by Hot coffee is equal to the thermal energy absorbed by ice cubes

So here we have

$Q_{ice} = Q_{coffee}$

now since ice cubes are added into coffee when it is at melting temperature

So here we can say that final temperature of coffee is T degree C

Now we have

$m_1L + m_1c_1\Delta T_1 = m_2c_2\Delta T_2$

here we have

$m_1 = 15 gram$

L = 333 kJ/kg = 333 J/g[/tex]

$c_1 = c_2 = 4186 J/kg C = 4.186 J/g C$

$\Delta T_1 = T - 0$

$\Delta T_2 = 85 - T$

now we have

$15(333) + 15(4.186)(T - 0) = 140(4.186)(85 - T)$

$4995 + 62.79T = 49813.4 - 586.04T$

$648.83 T = 44818.4$

$T = 69^o C$

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2 years ago

Which requires more work, lifting a 10.0kg load a vertical distance of 2m or lifting a 5.0kg load a distance of 4m?

Minchanka [31]

For each load, Work = (mass) x (gravity) x (distance .

Bigger load: Work = (10 kg) x (9.8 m/s²) x (2 m) = 196 joules .

Smaller load: Work = (5 kg) x (9.8 m/s²) x (4 m) = 196 joules.

The work required is equal in both cases.

The mass ratio of 2:1 is exactly balanced by
the height ratio of 1:2 .

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2 years ago

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Gnoma [55]

Two terminal device which can maintain a fixed voltage. Hope this help I need more detail in your question this is all I can provide. :)

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2 years ago