All categories

2 years ago

What does density have to do with heat?

2 answers:

8 0

If you take a fluid (i.e. air or water) and heat it, the portion that is heated usually expands. The same mass takes up more volume and as a consequence the heated portion becomes less dense than the portion that is not heated.

Vadim26 [7]2 years ago

7 0

Cooling a substance causes molecules to slow down and get slightly closer together, occupying a smaller volume that results in an increase in density. Hot water is less dense and will float on room-temperature water. Cold water is more dense and willsink in room-temperature water.

You might be interested in

A 70.0 kg astronaut is training for accelerations that he will experience upon reentry. He is placed in a centrifuge (r = 15.0 m

Levart [38]

Answer:

1.3823 rad/s

20.7345 m/s

28.66129935 m/s²

$a=2.92164g$

2006.29095 N radially outward

Explanation:

r = Radius = 15 m

m = Mass of person = 70 kg

g = Acceleration due to gravity = 9.81 m/s²

Angular velocity is given by

$\omega=13.2\times \dfrac{2\pi}{60}\\\Rightarrow \omega=1.3823\ rad/s$

Angular velocity is 1.3823 rad/s

Linear velocity is given by

$v=r\omega\\\Rightarrow v=15\times 1.3823\\\Rightarrow v=20.7345\ m/s$

The linear velocity is 20.7345 m/s

Centripetal acceleration is given by

$a_c=r\omega^2\\\Rightarrow a_c=15\times 1.3823^2\\\Rightarrow a_c=28.66129935\ m/s^2$

The centripetal acceleration is 28.66129935 m/s²

Acceleration in terms of g

$\dfrac{a}{g}=\dfrac{28.66129935}{9.81}\\\Rightarrow a=2.92164g$

$a=2.92164g$

Centripetal force is given by

$F_c=ma_c\\\Rightarrow F_c=70\times 28.66129935\\\Rightarrow F_c=2006.29095\ N$

The centripetal force is 2006.29095 N radially outward

The torque will be experienced when the centrifuge is speeding up of slowing down i.e., when it is accelerating and decelerating.

3 0

3 years ago

A region of space that contains no matter

sammy [17]

Conduction. Any material that easily allows heat to move through it. Vacuum. A region of space that contains no matter. Solid.

3 0

2 years ago

Read 2 more answers

An observer is standing next to the tracks, watching a train approach. The train travels at 30 m/s and blows its whistle at 8,00

SSSSS [86.1K]

7351.35Hz

f0= v-Vo/v-Vs × FSA

= 340-0 /340+30 ×8000

= 340/370× 8000

= 7351.35hz

7 0

3 years ago

A racecar driver circles a racetrack with a constant speed of 110 km/hr. Which BEST describes the speed and velocity of the car?

tangare [24]

The answer is B, The speed is constant and the velocity is changing.

8 0

2 years ago

Read 2 more answers

A 1.0 × 10^3 kg sports car is initially traveling at 15 m/s. The driver then applies the brakes for several seconds so that -25

algol13

Answer:

Ea = 112500[J]

Eb = 87500[J]

Explanation:

To solve this problem we must use the principle of energy conservation which tells us that the energy of a body plus the work done or applied by the body equals the final energy of a body.

This can be easily visualized by the following equation:

$E_{A}+E_{friction}=E_{B}$

Now we must define the energies at points A & B.

For point A

At point A we only have kinetic energy since it moves at 15 [m/s]

So the kinetic energy

$E_{A}=\frac{1}{2}*m*v_{A}^{2} \\E_{A}=\frac{1}{2} *1000*(15)^{2} \\E_{A}=112500[J]$

The final kinetic energy can be calculated as follows:

$112500-25000=E_{B}\\E_{B}=87500[J]$

8 0

2 years ago