All categories

3 years ago

Ethan is using a machine to mix cement. The work input is 10 and the work output is 8. What is the efficiency of the

1 answer:

4 0

Efficiency or $\eta$ is calculated by formula: $\eta=\frac{W_{output}}{W_{input}}$ where $\eta\leq1$ also efficiency has no unit.

Now put in the data and calculate the efficiency.

$\eta=\frac{8}{10}=\boxed{0.8}$

The efficiency of the machine is 0.8

If you want to get percentage just multiply by 100 and you get 80%

Hope this helps.

r3t40

You might be interested in

Is there a such thing as an elastic collision? Why do we study the case of the elastic collision in Physics?

fredd [130]

Yes there is an elastic collision in physics its when a collision occurs but no kinetic energy is loss. We study them in order to understand how to conserve momentum.

6 0

3 years ago

Which image shows both potential and kinetic energy?

Natasha2012 [34]

Answer:

The one in the middle

Explanation: i listened to the other person and i got it wrong, this is the answer for edge2020 sience review on energy!!!!

trust me its the middle one!!!!!

And everyone if ur not sure, like 100% sure about an answer dont answer at all cuz for 1: ur taking up a spot for others to answer. for 2: you could make people wrong. And for 3: its annoying. And 4: it makes stuff like this happen!

<u>NOT ARGUEING IM JUST PUTTING MY THOUGHTS AND OPINIONS OUT THERE ;)</u><em> many thanks.</em>

4 0

3 years ago

Read 2 more answers

A mother and daughter press their hands together and then push apart while ice skating. Immediately after they push away from ea

kherson [118]

Answer:

Explanation:

The daughter moves with greater acceleration backwards because of her weight.

7 0

3 years ago

Read 2 more answers

An electron in the first energy level of the electron cloud has an electron in the third energy level

lukranit [14]

Answer:

a lower energy than

Explanation:

3 0

3 years ago

Read 2 more answers

A loaded ore car has a mass of 950 kg. and rolls on rails ofnegligible friction. It starts from rest ans is pulled up a mineshaf

stiks02 [169]

(a) 10241 W

In this situation, the car is moving at constant speed: this means that its acceleration along the direction parallel to the slope is zero, and so the net force along this direction is also zero.

The equation of the forces along the parallel direction is:

$F - mg sin \theta = 0$

where

F is the force applied to pull the car

m = 950 kg is the mass of the car

$g=9.8 m/s^2$ is the acceleration of gravity

$\theta=30.0^{\circ}$ is the angle of the incline

Solving for F,

$F=mg sin \theta = (950)(9.8)(sin 30.0^{\circ})=4655 N$

Now we know that the car is moving at constant velocity of

v = 2.20 m/s

So we can find the power done by the motor during the constant speed phase as

$P=Fv = (4655)(2.20)=10241 W$

(b) 10624 W

The maximum power is provided during the phase of acceleration, because during this phase the force applied is maximum. The acceleration of the car can be found with the equation

$v=u+at$

where

v = 2.20 m/s is the final velocity

a is the acceleration

u = 0 is the initial velocity

t = 12.0 s is the time

Solving for a,

$a=\frac{v-u}{t}=\frac{2.20-0}{12.0}=0.183 m/s^2$

So now the equation of the forces along the direction parallel to the incline is

$F - mg sin \theta = ma$

And solving for F, we find the maximum force applied by the motor:

$F=ma+mgsin \theta =(950)(0.183)+(950)(9.8)(sin 30^{\circ})=4829 N$

The maximum power will be applied when the velocity is maximum, v = 2.20 m/s, and so it is:

$P=Fv=(4829)(2.20)=10624 W$

(c) $5.82\cdot 10^6 J$

Due to the law of conservation of energy, the total energy transferred out of the motor by work must be equal to the gravitational potential energy gained by the car.

The change in potential energy of the car is:

$\Delta U = mg \Delta h$