Question

(9). A machine has a mechanical advantage of 0.6. What force should be applied to the machine to make it apply 600 N to an object?

(2). Carlota does 2000 J of work on a machine. The machine does 500 J of work. What is the efficiency of the machine?

(4). Which best describes a difference between energy transformations in power plants and dams?

(5). A 40 kg dog is sitting on top of a hillside and has a potential energy of 1,568 J. What is the height of the hillside? (Formula: PE = mgh)

(6). What is the overall energy transformation in a coal-fired power plant?

(10). Which is an example of potential energy?

Answer 1

Step-by-step explanation:

First problem:

The first step you always need to do is to analyse the problem to find the proper theory and equation to solve it. In this case, the equation to calculate Mechanical Advantage is:
`MA=(of)/(if)`; where of means output force, and if means input force.

In this case, MA = 0.6; of = 600 N

So, replacing values into the equation:
`0.6=(600N)/(if) </p><p>[tex]if=(600N)/(0.6) =1000N`

This means that we need to applied a 360N force to have 600N over the object.

Second problem: It's about efficiency, which is defined as the ratio between the input and output work. It can be calculated with:
`eff=(W_(o) )/(W_(i) ) =(500J)/(2000J) = 0.25`;

`W_(o)=500J\\ W_(i)=2000J`

Which means that this machine has a 25% of efficiency.

Third problem: The difference is about the type of energy that is being transformed. In a power plant, the energy involved is chemical or electrical, nuclear power plants for example. On the other hand, dams involve only cinematic and potential energy to convert it into electric energy.

Fourth problem: This problem is about potential energy that involve gravity. The equation to calculate potential energy, includes height:
`PE=mgh`. In this case, m = 40kg, PE = 1,568 J.

Then, we replace variables and solve fro h:
`PE = mgh\\1,568 J =(40kg)(9.8m/s^(2))h`

`(1,568J)/((40kg)(9.8m/s^(2) )) =h\\h=(1,568)/(392) m\\h=4m`

Hence, the dog is at 4m.

Fifth problem: Energies involves in a coal-fired power is thermal, potential and kinetic to produce electrical energy.

Sixth problem: A daily life potential energy example is water tanks placed at specific heights. Due to its position from the ground, the gravity and water mass, an potential energy is present there.

Answer 2

(9) Mechanical advantage = force by machine / force applied to machine
0.6 = 600 / F
F = 1000 N

(2) Efficiency = (output / input) x 100
Efficiency = (500 / 2000) x 100
Efficiency = 25%

(4) The overall energy conversion in power plants is chemical to electrical while in dams it is potential to electrical.

(5) Using the formula:
1568 = 40 x 9.81 x h
h = 4.0 m

(6) Potential to electrical

(10) An object raised and held stationary above the ground.