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2 years ago

Is there a "real" simple machine that has an efficiency of 100%?

Physics

1 answer:

VARVARA [1.3K]2 years ago

8 0

Answer:

Explanation:

No, a machine cannot be 100% efficient. This is due to the movement of the moving parts siding against each other and causing friction. This friction is the one that creates heat and causes wear and tear between moving ports f the machine hence making the machine to decrease in efficiency with time

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An object has a mass of 10 kilograms and is accelerating at 5 m/s/s, what force pushed the object?

Katen [24]

Answer:

force pushed the object = 10 × 5 = 50N

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2 years ago

Help<br> How much current will flow when a 120 V power supply is connected to a 30<br> resistor ?

AVprozaik [17]

Current= voltage divided by resistance
120/30=4

7 0

2 years ago

1 point

Alecsey [184]

Answer:

Explanation:

ℎ

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2 years ago

What is the distance to a star whose parallex is 0.1 sec?

Arte-miy333 [17]

Answer:

$30.86\times 10^{13} km$

Explanation:

Given the parallex of the star is 0.1 sec.

The distance is inversely related with the parallex of the star. Mathematically,

$d=\frac{1}{P}$

Here, d is the distance to a star which is measured in parsecs, and P is the parallex which is measured in arc seconds.

Now,

$d=\frac{1}{0.1}\\d=10 parsec$

And also know that,

$1 parsec=3.086\times 10^{13} km$

Therefore the distance of the star is $30.86\times 10^{13} km$ away.

6 0

2 years ago

At what minimum speed must a roller coaster be traveling when upside down at the top of a circle so that the passengers do not f

worty [1.4K]

Answer:

$v_{min} \approx 17.153\,\frac{m}{s}$

Explanation:

The roller coaster begins with maximum kinetic energy and no gravitational potential energy. The gravitational potential energy reaches its maximum when roller coaster is upside down at the top of the circle. The physical model for the roller coaster is constructed by means of the Principle of Energy Conservation:

$\frac{1}{2}\cdot m \cdot v_{min}^{2} = m\cdot g \cdot h$

The minimum velocity is:

$v_{min} = \sqrt{2\cdot g \cdot h}$

Let assume that radio of curvature is measured in meters. Hence:

$v_{min} = \sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot(15\,m)}$

$v_{min} \approx 17.153\,\frac{m}{s}$

8 0

2 years ago