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2 years ago

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In a cyclic process, a gas performs 123 J of work on its surroundings per cycle. What amount of heat, if any, transfers into or

out of the gas per cycle?
123 J transfers out of the gas

123 J transfers into the gas

246 J transfers into the gas

0 J (no heat transfers)

1 answer:

Margaret [11]2 years ago

8 0

Answer:

123 J transfer into the gas

Explanation:

Here we know that 123 J work is done by the gas on its surrounding

So here gas is doing work against external forces

Now for cyclic process we know that

$\Delta U = 0$

so from 1st law of thermodynamics we have

$dQ = W + \Delta U$

$dQ = W$

so work done is same as the heat supplied to the system

So correct answer is

123 J transfer into the gas

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the physics concept corresponding to oomph is momentum. building on your above answers, figure out a formula for momentum (oomph

topjm [15]

Momentum = mass * velocity

I need parts A and B to explain the intuitions.

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2 years ago

Select all that apply to electrons and energy levels. - Scientists have not yet determined exactly why electrons do not collapse

Elenna [48]

Answer:

- The limitation of the maximum number of electrons in a given energy level can be used to account for the periodic recurrence of properties as the number of electrons increases.

Explanation:

First - Scientists have not yet determined exactly why electrons do not collapse into the nucleus. FALSE: Scientists do know why electrons do not collapse. Since the beginins of quantum mechanics it's known that the energy at small scales is quantized, that means there only can be certain values meaning that the energy do not change continously. In the case of the electron, it can only have certain levels of energy, that means they do not radiate continously as the go arround the atom, instead it is only allowed to have a certain amount of energy in a given state therefore it can not lose energy continously collapsing into the nucleus.

Second - Electrons cannot be located between levels except when they are in the process of moving. FALSE: We can not say that a electron moves between energy levels, it only can exist in any of the levels, but never in between. Also, the electron in any of its possible energy lavels can not be located with complete certainty due to the uncertainty principle.

Fourth - Electrons have any random energy. FALSE: as exposed above the electrons can only have certain cuantized energy levels acordinly to the rules of quantum physics

Fifth - Electrons can be found between energy levels. FALSE: Like said before we can not say that a electron exists between energy levels, it only can exist in any of the allowed levels, but never in between.

Thirth (correct one) : - The limitation of the maximum number of electrons in a given energy level can be used to account for the periodic recurrence of properties as the number of electrons increases. TRUE: the maximum number of electrons allowed in a given energy level directly determines the tipe of bond an atom can made with another (this due to the number of electrons in the higest energy level), so for example the elements in the left of a given row of the periodic table tend to have ionic bonds, but in the other hand the elements on right side tend form more covalent bonds. And this characteristic directly correllate with diferent properties of the elements.

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3 years ago

The motion of a particle is described by the position function s(t) = 2t - 15t +33t+17,t>0 , where is measured in seconds and

8090 [49]

The time when the particle is at rest is at 1.63 s or 3.36 s.

The velocity is positive at when the time of motion is at $0$ .

The total distance traveled in the first 10 seconds is 847 m.

<h3>When is a particle at rest?</h3>

A particle is at rest when the initial velocity of the particle is zero.

The time when the particle is at rest is calculated as follows;

s(t) = 2t³ - 15t² + 33t + 17

$v = \frac{ds}{dt} = 6t^2 -30t + 33\\\\at \ rest, \ v = 0\\\\6t^2 - 30t + 33 = 0\\\\6(t- \frac{5}{2} )^2- \frac{9}{2} = 0\\\\t = 1.63\ s \ \ or \ 3.36 \ s$

The velocity is positive at when the time of motion is as follows;

$0$ .

The total distance traveled in the first 10 seconds is calculated as follows;

$2(10)^3 - 15(10)^2 + 33(10) + 17 = 847 \ m$

Learn more about motion of particles here: brainly.com/question/11066673

4 0

2 years ago

A particle leaves the origin with a speed of 3 106 m/s at 38 degrees to the positive x axis. It moves in a uniform electric fiel

Salsk061 [2.6K]

Answer:

If the particle is an electron $E_y = 3.311 * 10^3 N/C$

If the particle is a proton, $E_y = 6.08 * 10^6 N/C$

Explanation:

Initial speed at the origin, $u = 3 * 10^6 m/s$

$\theta = 38^0$ to +ve x-axis

The particle crosses the x-axis at , x = 1.5 cm = 0.015 m

The particle can either be an electron or a proton:

Mass of an electron, $m_e = 9.1 * 10^{-31} kg$

Mass of a proton, $m_p = 1.67 * 10^{-27} kg$

The electric field intensity along the positive y axis $E_y$ , can be given by the formula:

$E_y = \frac{2 m u^2 sin \theta cos \theta}{qx} \\$

If the particle is an electron:

$E_y = \frac{2 m_e u^2 sin \theta cos \theta}{qx} \\$

$E_y = \frac{2 * 9.1 * 10^{-31} * (3*10^6)^2 *(sin38)( cos38)}{1.6*10^{-19} * 0.015} \\$

$E_y = 3311.13 N/C\\E_y = 3.311 * 10^3 N/C$

If the particle is a proton:

$E_y = \frac{2 m_p u^2 sin \theta cos \theta}{qx} \\$

$E_y = \frac{2 * 1.67 * 10^{-27} * (3*10^6)^2 *(sin38)( cos38)}{1.6*10^{-19} * 0.015} \\$

$E_y = 6.08 * 10^6 N/C$

8 0

2 years ago

At an accident scene on a level road, investigators measure a car's skid mark to be 84 m long. It was a rainy day and the coeffi

Flura [38]

The given data is incomplete. The complete question is as follows.

At an accident scene on a level road, investigators measure a car's skid mark to be 84 m long. It was a rainy day and the coefficient of friction was estimated to be 0.36. Use these data to determine the speed of the car when the driver slammed on (and locked) the brakes. (why does the car's mass not matter?)

Explanation:

Let us assume that v is the final velocity and u is the initial velocity of the car. Let s be the skid marks and $\mu$ be the friction coefficient and m be the mass of car.

Hence, the given data is as follows.

v = 0, s = 84 m, $\mu$ = 0.36

According to Newton's law of second motion the expression for acceleration is as follows.

F = ma

$-\mu N$ = ma

$-\mu mg$ = ma

a = $-\mu g$

Also,

$v^{2} = u^{2} + 2as$

$(0)^{2} = u^{2} + 2(-\mu g)s$

$u^{2} = 2(\mu g)s$

= $\sqrt{2(0.36)(9.81 m/s^{2})(84 m)}$

= 24.36 m/s

Thus, we can conclude that the speed of the car when the driver slammed on (and locked) the brakes is 24.36 m/s.

4 0

3 years ago