If the kinetic and potential energy in a system are equal, then the potential energy increases. What happens as a result?

What mass of water will fill a tank that is 100.0 cm long, 50.0 cm wide, and 30.0 cm high? Express the answer in grams.

Bond [772]

$calculate\ mass\ from\ formula\ for\ density:\\\\
density=volume*mass\\
mass=\frac{density}{volume}\\\\
volume=width*length*height\\
volume=100*50*30=150000cm^3=0,15m^3\\\\
density\ of\ water=\ 999.9720 \frac{kg}{m^3} \\\\mass=0,15*999.9720=149,9958kg=149995,8grams\\\\ Answer\ is\ :\\water\ which\ weight\ is\ equal\ to\ 149995,8grams.
$

4 0

3 years ago

A heat engine accepts 200,000 Btu of heat from a source at 1500 R and rejects 100,000 Btu of heat to a sink at 600 R. Calculate

diamong [38]

To solve the problem it is necessary to apply the concepts related to the conservation of energy through the heat transferred and the work done, as well as through the calculation of entropy due to heat and temperatra.

By definition we know that the change in entropy is given by

$\Delta S = \frac{Q}{T}$

Where,

Q = Heat transfer

T = Temperature

On the other hand we know that by conserving energy the work done in a system is equal to the change in heat transferred, that is

$W = Q_{source}-Q_{sink}$

According to the data given we have to,

$Q_{source} = 200000Btu$

$T_{source} = 1500R$

$Q_{sink} = 100000Btu$

$T_{sink} = 600R$

PART A) The total change in entropy, would be given by the changes that exist in the source and sink, that is

$\Delta S_{sink} = \frac{Q_{sink}}{T_{sink}}$

$\Delta S_{sink} = \frac{100000}{600}$

$\Delta S_{sink} = 166.67Btu/R$

On the other hand,

$\Delta S_{source} = \frac{Q_{source}}{T_{source}}$

$\Delta S_{source} = \frac{-200000}{1500}$

$\Delta S_{source} = -133.33Btu/R$

The total change of entropy would be,

$S = \Delta S_{source}+\Delta S_{sink}$

$S = -133.33+166.67$

$S = 33.34Btu/R$

Since $S\neq 0$ the heat engine is not reversible.

PART B)

Work done by heat engine is given by

$W=Q_{source}-Q_{sink}$

$W = 200000-100000$

$W = 100000 Btu$

Therefore the work in the system is 100000Btu

4 0

3 years ago

Please please please please PLEASE help!!!

inna [77]

Answer:

Because the electricity flows through and creates static bonds around the metal case which creates a bond with other fields that protects it.

Explanation:

8 0

2 years ago

A cat runs at 2 m/s for 3 s and then slows to a stop with an acceleration of 0.80 m/s2. What is

sp2606 [1]

Answer:

see that the correct one is B

Explanation:

To solve this exercise let us use the kinematic relations

v² = v₀² - 2 a x

as they indicate that the car stops, therefore the final speed is yield v = 0

x = v₀² / 2a

let's calculate

x = 2²/(2 0.8)

x = 2.5 m / s²

When reviewing the answers we see that the correct one is B

7 0

2 years ago

A 5 kg object near Earth's surface is released from rest such that it falls a distance of 10 m. After the object falls 10 m, it

makkiz [27]

Answer:D

Explanation:

Given

mass of object $m=5 kg$

Distance traveled $h=10 m$

velocity acquired $v=12 m/s$

conserving Energy at the moment when object start falling and when it gains 12 m/s velocity

Initial Energy $=mgh=5\times 9.8\times 10=490 J$

Final Energy $=\frac{1}{2}mv^2+W_{f}$

$=\frac{1}{2}\cdot 5\cdot 12^2+W_{f}$

where $W_{f}$ is friction work if any

$490=360+W_{f}$

$W_{f}=130 J$

Since Friction is Present therefore it is a case of Open system and net external Force is zero

An open system is a system where exchange of energy and mass is allowed and Friction is acting on the object shows that system is Open .

4 0

3 years ago