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## Two objects have masses of 39 MG and 21 MG. How much does the gravitational potential energy between the objects change if the distance between them changes from 8 m to 3 m?

Parzival S.
Featured 1 month ago

$\cong .0114 J$

#### Explanation:

Gravitational Potential Energy can be found with the equation:

$G P E = - \frac{{m}_{1} {m}_{2} G}{r}$

http://www.softschools.com/formulas/physics/potential_energy_twobody_gravitation_formula/36/

Therefore the change in GPE is:

$\Delta G P E = - \frac{{m}_{1} {m}_{2} G}{r} _ 2 - \left(- \frac{{m}_{1} {m}_{2} G}{r} _ 1\right)$

We can see that it will change linearly. The percent it will change by will be:

-1/3+1/8=-8/24+3/24=-5/24~=-20.83%

The actual change is (I'll interpret the units MG as Millions of Grams - and since the formula uses kilograms as the unit of mass, I'll adjust the numbers to get the units right):

$= - \frac{\left(39 \times {10}^{3}\right) \left(21 \times {10}^{3}\right)}{3} \left(6.673 \times {10}^{-} 11\right) + \frac{\left(39 \times {10}^{3}\right) \left(21 \times {10}^{3}\right)}{8} \left(6.673 \times {10}^{-} 11\right)$

$\cong - \frac{.0547}{3} - \frac{.0547}{8} \cong .0114 J$

How much energy is this?

One way to view it is to say that it's the energy required to move a paperclip 1 metre (3 feet).

~~~~~

The following is the original answer, which answered the question regarding gravitational force:

The force of gravity between two masses can be found with:

${F}_{g} = \frac{{m}_{1} {m}_{2}}{r} ^ 2 G$

Since we're changing the distance between the two objects, one way we can express the change is to see that:

${\left({r}_{1}\right)}^{2} = {8}^{2} = 64 = \frac{{m}_{1} {m}_{2}}{F} _ \left(g 1\right) G$

${\left({r}_{2}\right)}^{2} = {3}^{2} = 9 = \frac{{m}_{1} {m}_{2}}{F} _ \left(g 2\right) G$

And so the change in the two forces is an 64/9=711.bar1% increase.

The actual numerical change is (I'll interpret the units MG as Millions of Grams - and since the formula uses kilograms as the unit of mass, I'll adjust the numbers to get the units right):

${F}_{g 2} - {F}_{g 1} = \frac{{m}_{1} {m}_{2}}{{r}_{2}} ^ 2 G - \frac{{m}_{1} {m}_{2}}{{r}_{1}} ^ 2 G$

$= \frac{\left(39 \times {10}^{3}\right) \left(21 \times {10}^{3}\right)}{3} ^ 2 \left(6.673 \times {10}^{-} 11\right) - \frac{\left(39 \times {10}^{3}\right) \left(21 \times {10}^{3}\right)}{8} ^ 2 \left(6.673 \times {10}^{-} 11\right)$

$\cong \frac{.0547}{9} - \frac{.0547}{64} \cong .005 N$

This is a remarkably small amount of force. If you put two snickers bars in your hand (100g), the amount of force pushing down on your hand is roughly 1N. To achieve .005N, eat one of the snickers bars (you don't need it) and take the other one and cut it up into 250 equal pieces. Take one piece from that and put it in your hand. That is roughly .005N.

## Why is thermal energy not exploited more considering it is available and free?

Cosmic Defect
Featured 1 month ago

Second Law of Thermodynamics comes in way of utilising this resource. Efficiency of conversion is so low as to not justify the cost spent in building machineries for this purpose.

#### Explanation:

Of the various forms of energies, Mechanical energy and Electrical energy are of high quality whereas Thermal energy is of low quality.

While the Energy Conservation Principle (First Law of Thermodynamics) allows one to convert energy from one form to another, the Second Law of Thermodynamics places constraints on the efficiency of this conversion, especially from low quality forms to high quality forms.

The machines that convert thermal energy into mechanical/electrical energies are called Heat Engines. Heat Engines draw thermal energy from a heat reservoir at high temperature (${T}_{h}$), convert a fraction of it to mechanical energy and dump the remaining into a sink at lower temperature (${T}_{c}$). The Heat Engine with the maximum possible efficiency allowed by the laws of physics is called the Carnot Engine.

Second Law of thermodynamics clearly rules out a 100% conversion. The efficiency of the Carnot Engine is related to the temperatures of the heat reservoir (${T}_{h}$) and the heat sink (${T}_{c}$) as follows:
$\setminus {\eta}_{\text{carnot}} = 1 - {T}_{c} / {T}_{h} = \frac{\setminus \Delta T}{T} _ h$ ...... (1)
Looking at this equation it is clear that a 100% efficient engine would require either a heat reservoir of infinite temperature (${T}_{h} = \setminus \infty$) or a heat sink of zero kelvin (${T}_{c} = 0$ $K$ - ruled out by the third law of thermodynamics).

It is fair to ask why 100% why not be happy with what you get. Conventional heat engines (automobile engines and conventional nuclear plant turbines) have efficiencies in the range of 25% to 35%. Here the temperature difference between the heat source and heat sink is of the order of $1000$ $K$.

Now if you consider ocean as huge reservoir of thermal energy, the temperature difference between cold ocean bottom and hot ocean surface is of the order of few kelvins and the surface temperature is only about ${27}^{o}$ $C = 300$ $K$. This gives a Carnot efficiency of less than 1%. Now remember that Carnot Engines are ideal engines with theoretical maximum efficiency. Real engines will have efficiencies less than this. So it makes no sense to build heat engines of this poor efficiency. Power required to make and maintain these engines will be of of the same order as the power delivered by them.

$$        SORRY about overshooting the word limits.


## How did Millikan calculate the elementary charge using all those data points? I understand that they were all multiples of the elementary charge but I’m wondering how he took 100’s of data points and determined the common multiple.

Cosmic Defect
Featured 1 month ago

Millikan had data for the charges on the oil droplet which when sorted was expected to show a linear trend ${q}_{n} = e . n$. Slope of this straight line ($e$) is the fundamental unit of charge which is found using the least-square method.

#### Explanation:

The charges on the oil droplets are expected to be some integral multiple of a unknown fundamental unit of charge $e$. When the data on the charges in the oil droplets are sorted, we expect to see a relation of the form ${q}_{n} = e . n$.

If we plot ${q}_{n}$ Vs $n$ we expect to see a straight line whose slope would be this fundamental unit of charge ($e$). Since Millikan's data may have errors it may not be a perfect line but one can make a least-squares linear fit of the data.

Remember that all entries in the sequence $\left\{{q}_{n}\right\}$ may not be present. In fact it need not even start with ${q}_{1}$. The sequence interval $\setminus \Delta {q}_{n} = {q}_{n} - {q}_{n - 1} = e$, gives a rough estimate of $e$ that is used mainly to figure out $n$.

Millikan's own data is given below. The third column shows that the sequence interval, a rough estimate of $e$. It is clear that the first data (19.66) is roughly $4$ times the sequence interval (4.9). So it is obvious that ${q}_{1} , {q}_{2}$ and ${q}_{3}$ are missing. Similarly if you look at the ${16}^{t h}$ row you see that sequence interval suddenly jumps showing a missing ${q}_{15}$

Millikan's own data is reproduced below.
$n \setminus q \quad \setminus q \quad {q}_{n} \setminus q \quad \setminus q \quad \setminus \Delta {q}_{n}$
$\setminus \quad \setminus \quad \setminus \times {10}^{- 10} \setminus \times {10}^{- 10}$
$\setminus q \quad \setminus q \quad \left(e s u\right) \setminus q \quad \left(e s u\right)$
$01 \setminus q \quad \ldots \ldots \ldots . \setminus q \quad \ldots \ldots \ldots .$
$02 \setminus q \quad \ldots \ldots \ldots . \setminus q \quad \ldots \ldots \ldots .$
$03 \setminus q \quad \ldots \ldots \ldots . \setminus q \quad \ldots \ldots \ldots .$
$04 \setminus q \quad 19.66 \setminus q \quad \setminus \quad \ldots \ldots \ldots .$
$05 \setminus q \quad 24.60 \setminus q \quad \setminus \quad 4.94$
$06 \setminus q \quad 29.62 \setminus q \quad \setminus \quad 5.02$
$07 \setminus q \quad 34.47 \setminus q \quad \setminus \quad 4.91$
$08 \setminus q \quad 39.38 \setminus q \quad \setminus \quad 4.91$
$09 \setminus q \quad 44.42 \setminus q \quad \setminus \quad 5.04$
$10 \setminus q \quad 49.41 \setminus q \quad \setminus \quad 4.99$
$11 \setminus q \quad 53.91 \setminus q \quad \setminus \quad 4.50$
$12 \setminus q \quad 59.12 \setminus q \quad \setminus \quad 5.21$
$13 \setminus q \quad 63.68 \setminus q \quad \setminus \quad 4.56$
$14 \setminus q \quad 68.65 \setminus q \quad \setminus \quad 4.97$
$15 \setminus q \quad \ldots \ldots \ldots . \setminus q \quad \ldots \ldots \ldots .$
$16 \setminus q \quad 78.34 \setminus q \quad \setminus \quad 9.69$
$17 \setminus q \quad 83.22 \setminus q \quad \setminus \quad 4.88$

Millikan's Experiement Data Source:
http://physics.nyu.edu/~physlab/Classical%20and%20Quantum%20Wave%20Lab/Millikan%20oil%20drop%2002-06-2009.pdf

Least Squares Fit: When we have measured data $\left({x}_{i} , {y}_{i}\right)$ which we want to make a linear fit of the form, $y = a x$, the least squares method tries to minimise the square of the deviation from the fit value:
${S}^{2} = \setminus {\sum}_{i} {\left({y}_{i} - a {x}_{i}\right)}^{2} = \setminus {\sum}_{i} \left({y}_{i}^{2} + {a}^{2} {x}_{i}^{2} - 2 a {x}_{i} {y}_{i}\right)$
To find the $a$ that minimises $S$, differentiate $S$ with respect to $a$, set it to zero and solve for $a$.

$\setminus \frac{\setminus \partial {S}^{2}}{\setminus \partial a} = \setminus {\sum}_{i} 2 a {x}_{i}^{2} - 2 {x}_{i} {y}_{i} = 0$
$\setminus \overline{a} = \setminus \frac{\setminus {\sum}_{i} {x}_{i} {y}_{i}}{\setminus {\sum}_{i} {x}_{i}^{2}}$

Applying this to Millikan's data:

$\setminus \overline{e} = \setminus \frac{\setminus {\sum}_{n = 4}^{n = 17} n {q}_{n}}{\setminus {\sum}_{n = 4}^{n = 17} {n}^{2}} = \frac{7589 \setminus \times {10}^{- 10} \setminus \quad e s u}{1546}$
$\setminus \quad = 4.909 \setminus \times {10}^{- 10} \setminus \quad e s u$

Therefore the least-square fit value of the fundamental unit of charge for Millikan's data, written to 4 significant digits is:
$\setminus {\overline{e}}_{e s u} = 4.909 \setminus \times {10}^{- 10} \setminus \quad e s u$

Converting from $e s u$ to $\text{coulomb}$
$\setminus {\overline{e}}_{C} = \setminus {\overline{e}}_{e s u} \setminus \times \setminus \sqrt{4 \setminus \pi \setminus {\epsilon}_{0}} \setminus \times {10}^{- 6} = 1.637 \setminus \times {10}^{- 19} \setminus \quad C$

## An object vibrates in sipmple harmonic motion with time period 1.6 sec after passing the equilibrium position by 0.2 sec the speed of the object was 1.0m/s, what is the amplitude of the simple harmonic motion of the object? A)0.25m B)0.36m C)3.93m D)18.58

dk_ch
Featured 1 month ago

Option ( B)

#### Explanation:

Let the object $\textcolor{g r e e n}{P}$ executes SHM on Y-axis maintaining its equilibrium position at origin O . Given that the time period of oscillation is $T = 1.6$s. The angular velocity $\omega$ of the reference point $\textcolor{red}{R}$ associated with SHM undergoing uniform circular motion will be given by $\omega = \frac{2 \pi}{T} = \frac{2 \pi}{1.6} \text{rad/s}$.

Considering that time count is started when the object is at origin. So we can write the equation for displacement $x$ in t sec as

$\textcolor{red}{y = a \sin \omega t \ldots . . \left[1\right]}$ , where $a$ is the amplitude of vibration

Here origin is also its equilibrium position and It returns repeatedly to this position after each complete oscillation for 1.6s.

So the velocity $v$ of the object after $t$ sec will be obtained by differentiating the above equation w r to t

$\textcolor{b l u e}{v = \frac{\mathrm{dy}}{\mathrm{dt}} = a \omega \cos \omega t \ldots \ldots . \left[2\right]}$

Now it is given that the velocity of the object after passing the equilibrium position by $0.2$s is $1 m \text{/s}$

So we can insert $t = 0.2$s, $v = 1 m \text{/s}$ in [2] to find out $a$

$v = a \omega \cos \omega t$

$1 = a \times \frac{2 \pi}{1.6} \times \cos \left(\frac{2 \pi}{1.6} \times 0.2\right)$

$\implies 1 = a \times \frac{2 \pi}{1.6} \times \cos \left(\frac{\pi}{4}\right)$

$\implies 1 = a \times \frac{2 \pi}{1.6} \times \frac{1}{\sqrt{2}}$

$\implies a = \frac{0.8 \times \sqrt{2}}{\pi} \approx 0.36 m$

## Gauss' Law?

Cosmic Defect
Featured 1 month ago

#### Explanation:

The electric field inside a good conductor cannot be non-zero. If the electric field has to vanish inside a conductor, it must get polarized such that $- 126 \setminus \mu C$ of polar charge will get distributed at the cavity wall in such a way that the field lines due to the central charge will terminate on these polar charges.

But since the conductor is electrically neutral, each of these negative polar charges will have a positive charge of same magnitude associated with them. These would have drifted to the surface of the good conductor. The electric field will continue outside the conductor originating from these outer $+ 126 \setminus \mu C$.

Interior: Let $r$ be the distance of point ${P}_{1}$ from the central charge ($q = + 126 \setminus \mu C$). Construct an imaginary gaussian sphere centered on the central charge and of radius $\frac{r}{2}$, so that the point ${P}_{1}$ lies on its surface.

By Gauss's law the electric flux through this surface is -

\oint\quad vecE.dvecs = q_{"tot"}/\epsilon_0;

The electric field is spherically symmetric and so will have the same magnitude (${E}_{1}$) everywhere on the surface of this sphere. This integral then reduces to -

E_1 [4\pi(r/2)^2] = q_{"tot"}/\epsilon_0; \qquad E_1 = q/(\pi\epsilon_0r^2)

q=+126\muC; \qquad r=

Exterior: Point ${P}_{2}$ lies inside the good conductor. So the electric field at that point vanishes.

If we imagine a gaussian sphere centered on the central charge such that point ${P}_{2}$ lies on its surface, the total charge inside this sphere is zero. Because the total charge includes the central charge $q = + 126 \setminus \mu C$ and the polarized charge ${q}_{p}^{-} = - 126 \setminus \mu C$.

\oint\quad vecE.dvecs = q_{"tot"}/\epsilon_0; \qquad

E_2[4\pir_2^2] = (q+q_p^{-})/epsilon_0 = 0;
Therefore, $\setminus \quad {E}_{2} = 0$.

## What is the speed of the cube on its trajectory?

A08
Featured 2 weeks ago

At the point of contact of the cube with the cone, in the absence of friction two forces act on it.

1. Weight $= m g$ acting downwards.
2. Normal reaction $R$ at the point of contact perpendicular to the surface of cone.

Let $\theta$ be equal to half the angle of the vertex angle of the cone.

For circular motion required centripetal force $= \frac{m {v}^{2}}{r}$
where $m$ is mass of cube, $v$ its velocity in a horizontal circle of radius $r$.

This is provided by the horizontal component of normal reaction. We have the equation

$R \cos \theta = \frac{m {v}^{2}}{r}$ .....(1)

In the state of equilibrium vertical components of all the forces must be zero. Therefore, we get

$R \sin \theta - m g = 0$
$\implies R \sin \theta = m g$ .....(2)

Dividing (2) by (1) we get

$\tan \theta = \frac{r g}{v} ^ 2$

Solving for $v$ we get

$v = \sqrt{\frac{r g}{\tan} \theta}$

Inserting given values we get

$v = \sqrt{\frac{0.2 \times 9.81}{\tan} {55}^{\circ}}$
$v = 1.17 m {s}^{-} 1$, rounded to one decimal place.

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