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Black_prince [1.1K]
3 years ago
14

based on the law of conservation of energy. how can we reasonably improve a machines ability to do work? A.move the machine to a

different gravitational field B.increase the friction between its moving parts C. reduce the friction between its moving parts or D. redefine the machines system boundaries?
Physics
2 answers:
natima [27]3 years ago
8 0

The correct answer is

C. reduce the friction between its moving parts


In fact, by reducing the friction between the moving parts of the machine, it is possible to reduce the energy wasted due to this friction; therefore, more input energy is converted into useful work, and this will improve the efficiency of the machine.

erica [24]3 years ago
5 0

Answer: We can reasonably improve a machines ability to do work by reducing the friction between the moving parts of machine. The correct answer is C.

Explanation:

According to the law of conservation of energy, the energy can neither be created nor destroyed. The energy can be converted from one form to another.

The energy is wasted due to the friction. It reduces the efficiency of the machines to do the work. There will be energy loss due to the friction. By reducing the friction, More energy can be converted into the work.

Therefore, we can reasonably improve a machines ability to do work by reducing the friction between the moving parts of machine.

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Explanation:

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2 years ago
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KatRina [158]

Answer:

The value is  x =  45.99 \  Au

Explanation:

From the question we are told that

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Generally the average distance of the asteroid from the sun is mathematically represented as

            R = \sqrt[3]{ \frac{G M * T^2 }{4 \pi} }

Here M is the mass of the sun with a value  

        M  =  1.99*10^{30} \  kg

         G  is the gravitational constant with value  G  =  6.67 *10^{-11}  \  m^3 \cdot kg^{-1} \cdot  s^{-2}

           R = \sqrt[3]{ \frac{6.67 *10^{-11}  * 1.99*10^{30} * [5.55 *10^{9}]^2 }{4 * 3.142 } }

=>       R = 6.88 *10^{12} \  m

Generally

         1.496* 10^{11}  \  m  \to  1 Au (Astronomical \  unit )

So

          R = 6.88 *10^{12} \  m \ \ \ \ \to \ \   x \  Au

=>      x =  \frac{6.88 *10^{12}}{1.496 *10^{11}}

=>       x =  45.99 \  Au

       

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5 0
3 years ago
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Answer:

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Kaylis [27]
<span>Data:

</span>R = 5 Ω
U = 12 V
i = ?
<span>
Formula:

</span>U = R*i → i =  \frac{U}{R}<span>

Solving:

</span>i = \frac{U}{R}
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\boxed{i = 2.4A}

Answer:

<span>The current through the resistor is 2.4 Amperes</span><span>

</span>
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