explain how the efficiency of an ideal meachine compares with the efficiency of a real machine. For 25 POINTS

2 answers:

7 0

An ideal machine is 100% efficient, meaning it puts out all of the energy you put in (ie. no loss due to friction..) A real machine can never be 100% efficient. You must always put more energy Into a real machine than you get out of it.

Tcecarenko [31]2 years ago

5 0

The efficiency of such a machine is 100%. An ideal machine is one in which no part of the input energy to the machine gets wasted and the whole input energy is converted into the useful work. The efficiency of such a machine is 100%.
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Describe how a planet, solar system, galaxy, and the universe are related in terms of size.

Alexus [3.1K]

- The Solar System contains the sun and objects that orbit it, including the eight planets, comets, and asteroids, and the Galaxy contains about 100 billion stars, of which the sun is one, as well as large clouds of gas and dust. - The universe contains all physical matter and energy

So Therefore Universe Is the biggest terms of size because it contains all physical matter and energy hope it helps

7 0

2 years ago

A machine

In-s [12.5K]

Answer:

Power_input = 85.71 [W]

Explanation:

To be able to solve this problem we must first find the work done. Work is defined as the product of force by distance.

$W = F*d$

where:

W = work [J] (units of Joules)

F = force [N] (units of Newton)

d = distance [m]

We need to bear in mind that the force can be calculated by multiplying the mass by the gravity acceleration.

Now replacing:

$W = (80*10)*3\\W = 2400 [J]$

Power is defined as the work done over a certain time. In this way by means of the following formula, we can calculate the required power.

$P=\frac{W}{t}$

where:

P = power [W] (units of watts)

W = work [J]

t = time = 40 [s]

$P = 2400/40\\P = 60 [W]$

The calculated power is the required power. Now as we have the efficiency of the machine, we can calculate the power that is introduced, to be able to do that work.

$Effic=0.7\\Effic=P_{required}/P_{introduced}\\P_{introduced}=60/0.7\\P_{introduced}=85.71[W]$