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3 years ago

Use the law of conservation of energy to explain why the work output of a machine can never exceed the work input

2 answers:

7 0

The law of conservation of energy states that energy is not created nor destroyed, rather it is only transferred or transmitted from one form to another. Now, the work exerted which is the input is only transformed to move the object which is now becomes the output work. Some are used to move the object, some are converted to other forms. That is why it is never greater than the input energy. Thank you for your question. Please don't hesitate to ask in Brainly your queries.

abruzzese [7]3 years ago

4 0

Answer:

Explanation:

According to the law of energy conservation- energy of a closed system can neither be created nor be destroyed it can be only transformed. In a machine we get supply of input which get converted into different forms, we can not create any additional energy as input so it is possible that all the input energy converts into output( extreme case, when losses are zero) otherwise input energy coverts into many other form(heat, friction,etc) in the process of generating output. So output of a machine can never exceeds the input.

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Nataly [62]

Answer:

1.for the first one 100centimeters make 1 meter therefore

100cm-1m

340cm-x

340/100

=3.4

the answer is supposed to be 3.4, maybe there's a mistake with the question or the answer

2.the weight of a body is given by the formula

mass×gravity,in this case the mass is 75kg and the gravity is 9.8

weight=75×9.8

=735N

3.for this one the mass of a body is given by the formula

mass=weight/gravity

=420/9.8

=42.8kg

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3 years ago

A straight line is drawn on the surface of a 14-cm-radius turntable from the center to the perimeter. A bug crawls along this li

sleet_krkn [62]

Answer:

v = $\left[\begin{array}{c}0.66&0\end{array}\right]$ m/s

Explanation:

The position vector r of the bug with linear velocity v and angular velocity ω in the laboratory frame is given by:

$\overrightarrow{r}=vtcos(\omega t)\hat{x}+vtsin(\omega t)\hat{y}$

The velocity vector v is the first derivative of the position vector r with respect to time:

$\overrightarrow{v}=[vcos(\omega t)-\omega vtsin(\omega t)]\hat{x}+[vsin(\omega t)+\omega vtcos(\omega t)]\hat{y}$

The given values are:

$t=\frac{x}{v}=\frac{14}{3.8}=3.7 s$

$\omega=\frac{45\times2\pi}{60s}=4.7\frac{1}{s}$

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Andre45 [30]

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4 years ago

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3 years ago

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Arturiano [62]

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Explanation:

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3 years ago