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3 years ago

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

1 answer:

7 0

-- If the system is 'closed', then nothing ... including energy ... can get in or out, and the total energy inside has to be constant.

If half of the energy in the system starts out as potential energy and the rest starts out as kinetic, and then the potential energy increases, there's only one place the increase could have come from ... it could only have been converted from kinetic energy. So the kinetic energy in the system must decrease.

In fact, this isn't even a "result". The kinetic energy has to decrease before the potential energy can increase, because that's where the increase has to come from.

If the system is 'open', then energy can come in and go out. If the potential energy inside suddenly increases, we don't know where it came from, so we can't say anything about what happens to the system.

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Under what conditions does q, the heat evolved or absorbed by the system in a physical or chemical process, equal the change in

Paladinen [302]

Answer: When $w=-P\Delta V$

Explanation:

According to first law of thermodynamics, energy can neither be created nor be destroyed. It can only be transformed from one form to another.

$\Delta H=\Delta E+P\Delta V$

$\Delta E=q+w$

$\Delta H$ = change in enthalpy

P = pressure

$\Delta V$ = change in volume

$\Delta E$ =Change in internal energy

q = heat absorbed or released

w = work done on or by the system

$\Delta H=q+w+P\Delta V$

Thus for $\Delta E=q+w$ , $w=-P\Delta V$

The heat evolved or absorbed by the system in a physical or chemical process, equal the change in enthalpy of the system when $w=-P\Delta V$

8 0

4 years ago

4. Provide an example of an isolated energy system and explain how it could be changed to create an open energy system. You may

GarryVolchara [31]

Easy, take the top off your Thermos bottle filled with hot coffee. Assuming perfect insulation, that hot coffee is isolated from the environment; but when the top is opened the heat can now escape to that environment.

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

4 0

3 years ago

Read 2 more answers

Electrons are emitted from a surface when light of wavelength 500 nm is shone on the surface but electrons are not emitted for l

liberstina [14]

Explanation:

Given: $\lambda = 500\:\text{nm} = 5×10^{-7}\:\text{m}$

$\nu = \dfrac{c}{\lambda} = \dfrac{3×10^8\:\text{m/s}}{5×10^{-7}\:\text{m}}$

$\:\:\:\:\:= 6×10^{14}\:\text{Hz}$

The work function $\phi$ is then

$\phi = h\nu = (6.626×10^{-34}\:\text{J-s})(6×10^{14}\:\text{Hz})$

$\:\:\:\:\:\:\:= 3.98×10^{-19}\:\text{J}$

3 0

3 years ago

What are the standard units of length, mass, and volume in the metric system?

antoniya [11.8K]

Answer:

Meter, Gram and Liter.

Explanation:

In the metric system, the standard units for the below are;

Length - Meter

Mass - Gram

Volume - Liter.

8 0

3 years ago

Calculate the length of wire.

galben [10]

Answer:

L = 169.5 m

Explanation:

Using Ohm's Law:

V = IR

where,

V = Voltage = 1.5 V

I = Current = 10 mA = 0.01 A

R = Resistance = ?

Therefore,

1.5 V = (0.01 A)R

R = 150 Ω

But the resistance of a wire is given by the following formula:

$R = \frac{\rho L}{A}$

where,

ρ = resistivity = 1 x 10⁻⁶ Ω.m

L = length of wire = ?

A = cross-sectional area of wire = πr² = π(0.6 mm)² = π(0.6 x 10⁻³ m)²

A = 1.13 x 10⁻⁶ m²

Therefore,

$150\ \Omega = \frac{(1\ x\ 10^{-6}\ \Omega .m)L}{1.13\ x\ 10^{-6}\ m^2}\\\\L = \frac{150\ \Omega(1.13\ x\ 10^{-6}\ m^2)}{1\ x\ 10^{-6}\ \Omega .m}\\\\$

L = 169.5 m

7 0

3 years ago