What is the kinetic energy of a 620.0 kg roller coaster moving with a velocity of 9.00 m/s?

A heat engine extracts 42. 53 kj from the hot reservior and exhausts 17. 69 kj into the cold reservior. what is the work done?

ludmilkaskok [199]

The answer is 24.84kJ.

We apply the expression for the work done by the heat engine is,

$W=E_{i n}-E_{o u t}$ . Putting all given values in the equation we get the final answer.

What is heat engine?

A heat engine is a machine that uses heat to generate power. It draws heat from a reservoir, uses that heat to produce work, such as move a piston or lift weights, and then releases that heat energy into the sink.
We are given:The heat input is $E_{i n}=42.53 \mathrm{~kJ}$ . The heat output is $E_{o u t}=17.69 \mathrm{~kJ}$ .
The expression for the work done by the heat engine is, $W=E_{i n}-E_{o u t}$
Substituting the given values in the above expression, we will get $W =42.53 \mathrm{~kJ}-17.69 \mathrm{~kJ}$ =24.84kJ.
Thus, the work done by the heat engine is 24.84kJ.

To learn more about heat engine visit: brainly.com/question/15735984

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6 0

1 year ago

You are writing a report about space travel. Where would be the best place to find information? (2 points)

Harrizon [31]

Answer: Contact the National Aeronautics and Space Association

Explanation:

The National Aeronautics and Space Association is an independent organization in US which aims to enable a safe, secure, efficient space journey and launch of new satellites, spacecraft beyond the earth's orbit. It can give better information for the space travel already conducted by the astronauts. Thus will help in writing the report as it is the best place to find the information.

5 0

2 years ago

Read 2 more answers

If you travel at 3 m/s for 12 seconds, how far did you travel?

ella [17]

Answer:

V=3 m/s

t=12 seconds

S=?

S=V×t

S=3×12

S=36meters

So distance you travel is 36meters.

4 0

2 years ago

Read 2 more answers

A pendulum has 918 J of potential energy at the highest point of a swing. How much kinetic energy will it have at the bottom of

GREYUIT [131]

The kinetic energy at the bottom of the swing is also 918 J.

Assume the origin of the coordinate system to be at the lowest point of the pendulum's swing. A pendulum, when raised to the highest point has potential energy since it is raised to a height h above the origin. At the highest point, the pendulum's velocity becomes zero, hence it has no kinetic energy. Its energy at the highest point is wholly potential.

When the pendulum swings down from its highest position, it gains velocity. Hence a part of its potential energy begins to convert itself into kinetic energy. If no dissipative forces such as air resistance exist, then, the law of conservation of energy can be applied to the swing.

Under the action of conservative forces, the total mechanical energy of a system remains constant.This means that the sum of the potential and kinetic energies of a body remains constant.

When the pendulum reaches the lowest point of its swing, it is at the origin of the chosen coordinate system. Its vertical displacement from the origin is zero, hence its potential energy with respect to the origin is zero. Therefore the entire potential energy of 918 J should have been converted into kinetic energy, according to the law of conservation of energy.

Thus, the kinetic energy of the pendulum at the lowest point of its swing is equal to the potential energy it had at its highest point, which is equal to <u>918 J.</u>

8 0

3 years ago

A student sits on a rotating stool holding two 3.09-kg masses. When his arms are extended horizontally, the masses are 1.08 m fr

schepotkina [342]

Answer:

a

The New angular speed is $w_f = 2.034 rad/s$

b

The Kinetic energy before the masses are pulled in is $KE_i = 3.101 \ J$

c

The Kinetic energy after the masses are pulled in is $KE_f = 8.192 \ J$

Explanation:

From the we are told that masses are 1.08 m from the axis of rotation, this means that

The radius $r =1.08m$

The mass is $m = 3.09\ kg$

The angular speed $w = 0.770 \ rad/sec$

The moment of inertia of the system excluding the two mass $I = 3.25 \ kg \cdot m^2$

New radius $r_{new} = 0.34m$

Generally the conservation of angular momentum can be mathematical represented as

$w_f = [\frac{I_i}{I_f} ]w_i .....(1)$

Where $w_f$ is the final angular speed

$w_i$ is the initial angular speed

$I_i$ is the initial moment of inertia

$I_f$ is the final moment of inertia

Moment of inertia is mathematically represented as

$I = m r^2$

Where I is the moment of inertia

m is the mass

r is the radius

So the Initial moment of inertia is given as

$I_i = moment \ of \ inertia \ of\ the \ two \ mass \ + 3.25 \ kg \cdot m^2$

$I_i = 2m r^2 + 3.25$

The multiplication by is because we are considering two masses

$I_i = 2 [(3.09)(1.08)^2] +3.25 = 10.46 kg \cdot m^2$

So the final moment of inertia is given as

$I_f = 2[(3.09)(0.34)^2] +3.25 = 3.96 \ kg \cdot m^2$

Substituting these values into equation 1

$w_f = [\frac{10.46}{3.96} ] * 0.77 = 2.034 \ rad/sec$

Generally Kinetic energy is mathematically represented in term of moment of inertia as

$KE = \frac{1}{2} * I * w^2$

Now considering the kinetic energy before the masses are pulled in,

$KE_i = \frac{1}{2} * I_i * w^2_i$

The Moment of inertia would be $I_i = 10.46 \ Kg \cdot m^2$

The Angular speed would be $w_i = 0.77 \ rad/s$

Now substituting these value into the equation above

$KE_i = \frac{1}{2} * (10.46) * (0.770)^2 = 3.101 J$

Now considering the kinetic energy after the masses are pulled in,

$KE_f = \frac{1}{2} * I_f * w^2_f$

The Moment of inertia would be $I_f = 3.96 \ Kg \cdot m^2$

The Angular speed would be $w_f = 2.034 \ rad/s$

Now substituting these value into the equation above

$KE_f= \frac{1}{2} *(3.96)(2.034)^2$

$= 8.192J$

8 0

3 years ago