All categories

2 years ago

The ramp has an efficiency of 80%. If you perform 600 J of work, how much useful work does the ramp do

Physics

1 answer:

PIT_PIT [208]2 years ago

5 0

The ramp does 480J of useful work with an efficiency of 80% .

<h3>What is efficiency of work done ?</h3>

Efficiency is the ratio of the useful energy released by a system to the input energy .
Mathematically, efficiency of energy = out put energy/ input energy

<h3>What is the useful work done by the ramp having efficiency 80% and an input work done 600J?</h3>

The efficiency =output work done/ input work done
80% =output work done/ 600J
output work done =( 80×600)/100

=480J

Thus, we can conclude that the useful work done by the ramp is 480J.

Learn more about efficiency of energy here:

brainly.com/question/14280607

#SPJ1

You might be interested in

When a liquid is cooled, the kinetic energy of the particles . The force of attraction between the particles , the space between

anzhelika [568]

Answer:

For the first blank, the answer is decreases. For the second blank, the answer is increases. And finally for the third blank, the answer is decreases.

Explanation:

For the first blank, the answer is decreases. For the second blank, the answer is increases. And finally for the third blank, the answer is decreases.

When a liquid is cooled, the kinetic energy of the particles decreases. The force of attraction between the particles increases, the space between the particles decreases, and the matter changes its state to solid.

4 0

3 years ago

Read 2 more answers

HEY CAN YALL PLS HELP ME IN DIS!!!!!!!!!!!!!

Serga [27]

Answer:

Hey!!

Your answer is: 0.72

Explanation:

if 760=1 then...

550=x

=550÷760= 0.72 in two s.f

6 0

3 years ago

You pull with a force of 295 N on a rope that is attached to a block of mass 22 kg, and the block slides across the floor at a c

Sergeeva-Olga [200]

Answer:

Fnet = 0

Explanation:

Since the block slides across the floor at constant speed, this means that it's not accelerated.
According Newton's 2nd Law, if the acceleration is zero, the net force on the sliding mass must be zero.
This means that there must be a friction force opposing to the horizontal component of the applied force, equal in magnitude to it:

$F_{appx} = F_{app} * cos \theta = 295 N * cos 35 = 242 N (1)$

In the vertical direction, the block is not accelerated either, so the sum of the normal force and the vertical component of the applied force, must be equal in magnitude to the force of gravity on the block:

$F_{appy} = F_{app} * cos \theta = 295 N * sin 35 = 169 N (2)$

⇒ 169 N + Fn = Fg = 216 N (3)

This means that there must be a normal force equal to the difference between Fappy and Fg, as follows:
Fn = 216 N - 169 N = 47 N (4)

6 0

3 years ago

If the ball shown in the figure lands in 0.5 s, about what height was it thrown from?

Ede4ka [16]

Answer: 1.22 m

Explanation:

The equation of motion in this situation is:

$y=y_{o}+V_{oy}t-\frac{g}{2}t^{2}$ (1)

Where:

$y=0$ is the final height of the ball

$y_{o}=h$ is the initial height of the ball

$V_{oy}=V_{o}sin(0\°)=0$ is the vertical component of the initial velocity (assuming the ball was thrown vertically and there is no horizontal velocity)