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Doss [256]
3 years ago
7

What is the energy put into a machine or device to make it work

Physics
1 answer:
Neporo4naja [7]3 years ago
4 0

Answer:

  • <em>The energy put into a machine or device to make it work is named</em> <u>work input.</u>

Explanation:

<em>Machines </em>are devices to perform tasks with less effort or more easily. They can either increase the force applied or change the direction of the force or permit to apply the force at a distance.

The law of conservation of energy states that the energy is neither created nor destroyed, but when the machines do <em>work</em> some energy is dissipated in the form of heat, light or sound, due, mainly to friction.

In consequence, the <em>work output</em> is less than the <em>work input</em>.

A measure of energy use is efficiency. The efficiency is the ratio of the work output to the work input.

          Efficiency=\dfrac{\text{Work done by the machine}}{\text{Work done on the machine}}

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3 years ago
A car with speed v and an identical car with speed 2v both travel the same circular section of an unbanked road. If the friction
yawa3891 [41]

Answer:

F'=\dfrac{F}{4}

Explanation:

Let m is the mass of both cars. The first car is moving with speed v and the other car is moving with speed 2v. The only force acting on both cars is the centripetal force.

For faster car on the road,

F=\dfrac{mv^2}{r}

v = 2v

F=\dfrac{m(2v)^2}{r}

F=4\dfrac{m(v)^2}{r}..........(1)

For the slower car on the road,

F'=\dfrac{mv^2}{r}............(2)

Equation (1) becomes,

F=4F'

F'=\dfrac{F}{4}

So, the frictional force required to keep the slower car on the road without skidding is one fourth of the faster car.

8 0
4 years ago
A parallel-plate vacuum capacitor has 8.38 J of energy stored in it. The separation between the plates is 2.30 mm. If the separa
Elanso [62]

Answer:

Explanation:

plate separation = 2.3 x 10⁻³ m

capacity C₁ = ε A / d

= ε A / 2.3 x 10⁻³

C₂ = ε A / 1.15 x 10⁻³

\frac{C_2}{C_1} = \frac{2.3}{1.15}

a ) when charge remains constant

energy = \frac{q^2}{2C}

q is charge and C is capacity

energy stored initially E₁= \frac{q^2}{2C_1}

energy stored finally E₂ = \frac{q^2}{2C_2}

\frac{E_1}{E_2} = \frac{C_2}{C_1} = \frac{2.3}{1.15}

E_2 = \frac{1.15}{2.3 } \times E_1

= \frac{1.15}{2.3 } \times 8.38

= 4.19 J

b )

In this case potential diff remains constant

energy of capacitor = 1/2 C V²

energy is proportional to capacity as V is constant .

\frac{E_2}{E_1} = \frac{C_2}{C_1}

\frac{E_2}{8.38} = \frac{2.3}{1.15}

E_2 = 16.76 .

8 0
3 years ago
Que es la accion y reaccion
alina1380 [7]

Answer:

what is the action and reaction ?

Explanation:

   

4 0
3 years ago
Please HELPPPPPP THIS IS TIMED
Troyanec [42]

Answer:

When a person runs, their body must convert potential energy into kinetic energy. Potential energy is the energy stored within a system.When a person runs, their body must convert potential energy into kinetic energy. Potential energy is the energy stored within a system. Potential energy is used when the system uses kinetic energy to move in a horizontal direction.

In the human body, potential energy is stored in the form of chemical energy. The chemical energy comes from the food that a person consumes throughout the day. The body needs a certain amount of calories (the energy from food) in order to perform certain activities. If the runner has not consumed enough calories throughout their day, they will run out of potential energy and become tired. This is because the body is not very efficient at retaining energy. Energy cannot be created or destroyed, but it can go elsewhere. As the person runs, most of their stored energy is released in the form of thermal energy. This is why people get hot and start to sweat when they do physical activity such as running. The body is heating up because it is literally burning the calories that it has consumed in order to keep moving in a horizontal direction. The runner will sweat, because sweating is the body's natural cooling mechanism. If the runner did not have the ability to sweat, the conversion of potential energy (the chemical energy) into kinetic energy which is released as thermal energy would cause the runner's body to overheat. The chemical energy that the runner consumes in the form of calories is also released in the form of sound energy. Every time the runner's foot hits the ground, energy is leaving the runner's body as sound waves emit from the impact of the runner's foot on the ground. Because energy is being released from the runner's body with every step they take, it is important for the runner to consume enough chemical energy in the form of calories prior to their run. The runner's body needs a substantial amount of calories as a reserve so that they will have more to burn as their potential energy is released throughout the run.

Thermal energy is measured in calories. Calories are released from a given item as it burns. The amount of calories that are in something depends directly on the amount of chemical bonds that are broken and formed as it burns. For example, when a piece of wood burns, 3000 calories of thermal energy are released per gram. When an apple is burned however, it releases about 600 calories of thermal energy. Therefore, it is reasonable to assume that there is more energy available from breaking the atomic bonds in wood than from breaking the atomic bonds in an apple.

One calorie is defined as the amount of thermal energy needed to raise the temperature of one gram of water one degree Celsius. Calories burn very slowly in the human body, and as they do, kinetic energy becomes available to the runner. 1 calorie is the equivalent to 4.186 Joules of energy. So, the more calories that the runner consumes prior to running, the more energy they will have available to them throughout the run. The runner�s energy can also be measured in the form of watts, or electrical power. One calorie also translates to about 4.186 watts. So, if the runner has 500 calories available to them, they are capable of producing over 2000 watts of electrical power.Kinetic energy is equal to one half of the runner's mass times their velocity squared (KE=1/2mv^2). So, if the runner has a mass of 60 kg and wants to run at a rate of 9m/s, they will use about 2,430 Joules of energy. The runner is not able to change their mass, but they can increase or decrease their use of kinetic energy by increasing or decreasing their velocity. If the runner has not consumed a lot of chemical energy throughout their day, it would be wise for them to decrease their velocity as to decrease their kinetic energy and therefore use less of their stored potential energy.

4 0
2 years ago
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