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3 years ago

Distinguish between thermal energy, temperature, and heat. (BE SPECIFIC & use physics vocabulary)

1 answer:

3 0

Thermal energy can be defined as specific thermodynamic quantities as a part of energy transfer such as heat, which is a form of thermal energy flowing from hotter to colder objects. Temperature is a physical property of an object which specify whether the subject is hot or cold.

Explanation:

Difference between thermal energy, temperature and heat

Thermal Energy

Thermal energy is the total kinetic energy of molecules bundled up in the object. This energy is the heat exerted by the object which forms its temperature.

Temperature

In technical terms, temperature is the average kinetic energies measurement of molecules and atoms, present in an object. Simply, we can say, temperature defines how hot or how cold is a substance. Temperature relies on the kinetic energy of particles.

Heat

Transfer of thermal energy from a hotter to colder objects. Heat is different from temperature. Temperature is the mark to define how hot a substance is whereas, heat is the flow of thermal energy.

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(TIMED) Anyone know the answer to this question?

KengaRu [80]

I believe C is your answer (also, I am also using tht same exact program!)

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3 years ago

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Letícia leaves the grocery store and walks 150.0 m to the parking lot. Then, she turns 90° to the right and walks an additional

Alexxx [7]

Answer:

165.529454

Explanation:

According to the Pythagorean Theorem for calculating the lengths of a right angle triangle's sides, a^2 + b+2 = c^2, where c is the longest side (and the side opposing the right angle). So in your case it would be 150*150 + 70*70 = 27400. And √ 27400 is your answer.

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2 years ago

A charge of 25 nC is uniformly distributed along a straight rod of length 3.0 m that is bent into a circular arc with a radius o

Greeley [361]

Answer:

E = 31.329 N/C.

Explanation:

The differential electric field $dE$ at the center of curvature of the arc is

$dE = k\dfrac{dQ}{r^2}cos(\theta )$ (we have a cosine because vertical components cancel, leaving only horizontal cosine components of E. )

where $r$ is the radius of curvature.

Now

$dQ = \lambda rd\theta$ ,

where $\lambda$ is the charge per unit length, and it has the value

$\lambda = \dfrac{25*10^{-9}C}{3.0m} = 8.3*10^{-9}C/m.$

Thus, the electric field at the center of the curvature of the arc is:

$E = \int_{\theta_1}^{\theta_2} k\dfrac{\lambda rd\theta }{r^2} cos(\theta)$

$E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2}cos(\theta) d\theta.$

Now, we find $\theta_1$ and $\theta_2$ . To do this we ask ourselves what fraction is the arc length 3.0 of the circumference of the circle:

$fraction = \dfrac{3.0m}{2\pi (2.3m)} = 0.2076$

and this is

$0.2076*2\pi =1.304$ radians.

Therefore,

$E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2} cos(\theta)d\theta= \dfrac{\lambda k}{r} \int_{0}^{1.304}cos(\theta) d\theta.$

evaluating the integral, and putting in the numerical values we get:

$E = \dfrac{8.3*10^{-9} *9*10^9}{2.3} *(sin(1.304)-sin(0))\\$

$\boxed{ E = 31.329N/C.}$

4 0

3 years ago

The gas pressure inside a container decreases when

avanturin [10]

Answer:

When the volume increases or when the temperature decreases

Explanation:

The ideal gas equation states that:

$pV= nRT$

where

p is the gas pressure

V is the volume

n is the number of moles of gas

R is the gas constant

T is the gas temperature

Assuming that we have a fixed amount of gas, so n is constant, we can rewrite the equation as

$\frac{pV}{T}=const.$

which means the following:

- Pressure is inversely proportional to the volume: this means that the pressure decreases when the volume increases

- Pressure is directly proportional to the temperature: this means that the pressure decreases when the temperature decreases

8 0

3 years ago

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 47.0 nC placed between q1

lesya [120]

Answer:

Incomplete question, check attachment for completed question

Explanation:

The force of attraction between two forces are given as

F=kQq/r²

4 0

3 years ago