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3 years ago

The first law of thermodynamics can be given as ________.

1 answer:

7 0

The first law of thermodynamics can be written as
$\Delta U = Q-W$
where
$\Delta U$ is the variation of internal energy of the system
$Q$ is the amount of heat absorbed by the system
$W$ is the work done by the system on the surrounding.

Using this form, the sign convention for Q and W becomes:
Q > 0 --> heat absorbed by the system (because it increases the internal energy)
Q < 0 --> heat released by the system (because it decreases the internal energy)
W > 0 --> work done by the system (for instance, an expansion: when the system expands, it does work on the surrounding, and so the internal energy decreases, this is why there is a negative sign in the formula Q-W)
W < 0 --> work done by the surrounding on the system (for instance, a compression: when the system is compressed, the surrounding is doing work on the system, and so the internal energy of the system increases)

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A4.0 kg object is moving with speed 2.0 m/s. A 1.0 kg object is moving with speed 4.0 m/s. Both objects encounter the same const

Evgesh-ka [11]

Answer:

Both objects travel the same distance.

Explanation:

Given that,

Mass of first object = 4.0 kg

Speed of first object = 2.0 m/s

Mass of second object = 1.0 kg

Speed of second object = 4.0 m/s

We need to calculate the stopping distance

For first particle

Using equation of motion

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

Where, v = final velocity

u = initial velocity

s = distance

Put the value in the equation

$0= u^2-2as_{1}$

$s_{1}=\dfrac{u^2}{2a}$ ....(I)

Using newton law

$a=\dfrac{F}{m}$

Now, put the value of a in equation (I)

$s_{1}=\dfrac{8}{F}$

Now, For second object

Using equation of motion

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

Put the value in the equation

$0= u^2-2as_{2}$

$s_{2}=\dfrac{u^2}{2a}$ ....(I)

Using newton law

$F = ma$

$a=\dfrac{F}{m}$

Now, put the value of a in equation (I)

$s_{2}=\dfrac{8}{F}$

Hence, Both objects travel the same distance.

6 0

3 years ago

Write the equation in slope-intercept form x-2y=4

maxonik [38]

Answer:

$y=\frac{1}{2}x-2$

Explanation:

Slope-intercept form means we want the y to be by itself in the equation. Every thing we do will be about getting the y alone on the left side of the equation

To start we should move x to the left hand side. We can do this by subtracting x from both sides. That way, there is an x on the right, but not the left.

x-x-2y=4-x

this gives us

-2y=4-x

Great! So now what? Well, the y isn't by itself yet because it still is being multiplied by negative two (-2). In order to move it from the left side to the right side, we have to do the opposite of multiply; divide. So, we will divide both sides by -2

$\frac{-2y}{-2} =\frac{4}{-2} -\frac{x}{-2}$

-2 divided by -2 is 1, 4 divided by -2 is -2, and -x divided by -2 is $\frac{1}{2}x$

This gives us the answer: $y=\frac{1}{2}x-2$

Tips:

A negative divided by a negative is a positive ex: -4 divided by -2 is positive 2

If you are subtracting by a negative number, you are actually adding by a positive ex: 2-(-2) is actually 2+2

Don't be afraid to have fractions in your equations

Whatever you do to the one side of the equation, you must do it to the other side as well. Multiply the left side by 2? You HAVE TO multiply the right side by two as well. Add 3 to the right side? You HAVE TO add 3 to the left.

For problems like this (and when you have access to the internet), where you need to rewrite an equation, double check your work on desmos, which is an online graphing calculator. Input both the original equation and the equation you rewrote, and if they don't create the same line, you did something wrong.

8 0

3 years ago

what is the rotational kinetic energy of the earth? use the moment of inertia you calculated in part a rather than the actual mo

Ivenika [448]

The Earth's rotational kinetic energy is the kinetic Energy that the Earth

has due to rotation.

The rotational kinetic energy of the Earth is approximately 3.331 × 10³⁶ J

Reasons:

The parameters required for the question are;

Mass of the Earth, M = 5.97 × 10²⁴ kg

Radius of the Earth, R = 6.38 × 10⁶ m

The rotational period of the Earth, T = 24.0 hrs.

$The \ moment \ of \ inertia \ of \ uniform \ sphere \ is \ I = \mathbf{\dfrac{2}{5} \cdot M \cdot R^2}$

Which gives;

$\mathbf{I_{Earth}} = \dfrac{2}{5} \times 5.97 \times 10 ^{24} \cdot \left(6.38 \times 10^6 \right)^2 = 9.7202107 \times 10^{37}$

$\mathrm{The \ rotational \ kinetic \ energy \ is} \ E_{rotational} = \mathbf{\dfrac{1}{2} \cdot I \cdot \omega^2}$

$\mathrm{The \ angular \ speed, \ \omega} = \mathbf{\dfrac{2 \dcdot \pi}{T}}$

Therefore;

$\omega = \dfrac{2 \cdot \pi}{24} = \dfrac{\pi}{24}$

Which gives;

$\mathbf{E_{rotational}} = \dfrac{1}{2} \times 9.7202107 \times 10^{37} \times \left( \dfrac{\pi}{12} \right)^2 = 3.331 \times 10^{36}$

The rotational kinetic energy of the Earth, $E_{rotational}$ = 3.331 × 10³⁶ Joules

Learn more here:

brainly.com/question/13623190

The moment of inertia from part A of the question (obtained online) is that of the Earth approximated to a perfect sphere.

Mass of the Earth, M = 5.97 × 10²⁴ kg

Radius of the Earth, R = 6.38 × 10⁶ m

The rotational period of the Earth, T = 24.0 hrs

3 0

3 years ago

Light takes 8 minutes to reach the Earth, and the speed of light is 3.0×10^8 m/s. a) What is the orbital speed of the Earth arou

spin [16.1K]

Answer:

(a) 28690 m/s (b) $2.46x10^{33}J$

Explanation:

The orbital speed is define as:

$v = \frac{2 \pi r}{T}$ (1)

Where r is the radius of the trajectory and T is the orbital period.

To determine the orbital speed of the Earth it is necessary to know the orbital period and the radius of the trajectory. That can be done by means of the Kepler's third law and average velocity equation.

The average velocity in a Uniform Rectilinear Motion is defined as:

$v = \frac{d}{t}$ (2)

Where v is the velocity, d is the covered distance and t is the time.

Equation 2 can be rewritten for d to get:

$d = vt$ (3)

In this case, v will be the speed of light and t, the 8 minutes that takes to reach the Earth.

The time will be converted to seconds so the units in equation 3 can match:

$8min . \frac{60s}{1min}$ ⇒ $480s$

$t = 480s$

Replacing all those values in equation 3 it is gotten:

$d = (3.0x10^{8}m/s)(480s)$

$d = 1.44x10^{11}m$

Kepler’s third law is defined as:

$T^{2} = r^{3}$

Where T is orbital period and r is the radius of the trajectory.

$T = \sqrt{r^{3}}$

$T = \sqrt{(1.44x10^{11}m)^{3}}$

It is necessary to pass from meters to astronomical unit (AU), 1 AU is defined as the distance between the Earth and the Sun.

$T = \sqrt{1AU}$

$T = 1AU$

That can be expressed in units of years.

$T = 1AU . \frac{1year}{1AU}$

$T = 1year$

But there are 31536000 seconds in one year:

$T = 1year . \frac{31536000s}{1year}$

$T = 31536000s$

Finally, equation 1 can be used:

$v = \frac{2 \pi (1.44x10^{11}m)}{(31536000s)}$

$v = 28690 m/s$

So Earth orbital speed around the Sun is 28690 m/s.

b) What is its kinetic energy?

The kinetic energy is defined as:

$E = \frac{1}{2}mv^{2}$ (4)

Notice that it is necessary to found the mass of the Earth, that can be done combining the Universal law of gravity and Newton's second law:

$F = \frac{GMm}{r^{2}}$

$ma = \frac{GMm}{r^{2}}$ (5)

M will be isolated in equation 5:

$M = \frac{r^{2}a}{G}$

Where r is the radius of the Earth ( $6.38x10^{6}m$ )

$M = \frac{(6.38x10^{6}m)^{2}(9.8m/s^{2})}{6.67x10^{-11}kg.m/s^{2}.m^{2}/Kg^{2}})$

$M = 5.98x10^{24} Kg$

$E = \frac{1}{2}( 5.98x10^{24} Kg))(28.690m/s)^{2}$

$E = 2.46x10^{33}Kg.m^{2}/s^{2}$

$E = 2.46x10^{33}J$

Hence, the kinetic energy of Earth is $2.46x10^{33}J$ .

8 0

3 years ago

A rock is rolling down a hill. At position 1, it’s velocity is 2.0 m/s. Twelve seconds later, as it passes position 2, it’s velo

mr Goodwill [35]

Answer

Hi,

correct answer is {D} 3.5 m/s²

Explanation

Acceleration is the rate of change of velocity with time. Acceleration can occur when a moving body is speeding up, slowing down or changing direction.

Acceleration is calculated by the equation =change in velocity/change in time

a= {velocity final-velocity initial}/(change in time)

a=v-u/Δt

The units for acceleration is meters per second square m/s²

In this example, initial velocity =2.0m/s⇒u

Final velocity=44.0m/s⇒v

Time taken for change in velocity=12 s⇒Δt

a= (44-2)/12 = 42/12

3.5 m/s²

Best Wishes!

5 0

3 years ago