# Make the internet a better place to learn

## Question bfcde

Zack M.
8.986301369863014 years ago

Yes, there is gravity in space, otherwise, all of our satellites around the Earth would float away. For that matter, Earth and the other planets would fly away from the Sun!!

#### Explanation:

Think about how gravity works. Objects are pulled to the center of the Earth. It's important to note its the center of the Earth, and not just "down." Now picture a cannon at the top of a huge tower. A cannon ball that is dropped from the tower will simply fall toward the base of the tower.

The cannon ball falls at a constant acceleration of 1g toward the earth, which Galileo showed is the same acceleration as everything else on the Earth! This is important to remember!!

What if the cannon ball is fired from the cannon, though? It still accelerates toward the earth at the same rate as the dropped ball, but it will move forward a bit while it falls. Because the Earth is round, the ground actually drops away from the ball! So the faster the ball is launched, the longer it will take to hit the ground.

In fact, if the ball was launched forward fast enough, it would never hit the ground at all. It would travel all the way around the earth and collide with the back of the cannon! This is how an orbit is achieved! Remember though, gravity is still pulling on the cannon ball. If it wasn't, the ball would just fly off into space in a straight line. That is Newton's first law, inertia, in action.

So the moon is like the cannon ball. It just moves forward as fast as it falls and therefore stays in orbit. As for the astronauts, they float in their space stations because the astronaut and the station are both falling toward the earth at the same rate. They are both in free fall.

The result is that the astronaut is never able to catch up to his spaceship. This is why he floats. You can experience this yourself with a trampoline and a baseball. As you bounce, release the baseball and from your perspective, it will appear to "rise up" from you hand, at least until you bounce again.

## Question de828

Olivier B.
8.87945205479452 years ago

The distance between the earth and the body must be 259 358 400 m for the gravitational attraction of the sun and the earth on the body being balanced.

#### Explanation:

The gravitational attraction between two bodies is calculated as follows:

$F = \frac{G \cdot {m}_{1} \cdot {m}_{2}}{d} ^ 2$

with $G$ the universal gravity constant, ${m}_{1}$ and ${m}_{2}$ the masses of the two bodies and $d$ the distance between the two bodies.

If we place a body on a straight line between the earth and the sun, the resulting gravitational attractions will be:

${F}_{s b} = \frac{G \cdot {m}_{s} \cdot {m}_{b}}{{\left({d}_{s b}\right)}^{2}}$

${F}_{e b} = \frac{G \cdot {m}_{e} \cdot {m}_{b}}{{\left({d}_{e b}\right)}^{2}}$

We are considering the situation when ${F}_{s b} = {F}_{e b}$:

$\frac{\cancel{G} \cdot {m}_{s} \cdot \cancel{{m}_{b}}}{{\left({d}_{s b}\right)}^{2}} = \frac{\cancel{G} \cdot {m}_{e} \cdot \cancel{{m}_{b}}}{{\left({d}_{e b}\right)}^{2}}$

$\rightarrow \frac{{m}_{s}}{{\left({d}_{s b}\right)}^{2}} = \frac{{m}_{e}}{{\left({d}_{e b}\right)}^{2}}$

$\rightarrow \frac{{m}_{s}}{{m}_{e}} = \frac{{\left({d}_{s b}\right)}^{2}}{{\left({d}_{e b}\right)}^{2}} = {\left(\frac{{d}_{s b}}{{d}_{e b}}\right)}^{2}$

$\rightarrow \sqrt{\frac{{m}_{s}}{{m}_{e}}} = \frac{{d}_{s b}}{{d}_{e b}}$

Knowing that ${m}_{e} = 6 \cdot {10}^{24}$kg and ${m}_{s} = 2 \cdot {10}^{30}$kg, and that ${d}_{s b} + {d}_{e b} = 1.5 \cdot {10}^{11}$m we have to solve the following equation:

$\sqrt{\frac{2 \cdot {10}^{30}}{6 \cdot {10}^{24}}} = \frac{1.5 \cdot {10}^{11} - {d}_{e b}}{{d}_{e b}}$

$\rightarrow {d}_{e b} = \frac{1.5 \cdot {10}^{11} - {d}_{e b}}{\sqrt{\frac{2 \cdot {10}^{30}}{6 \cdot {10}^{24}}}} = \frac{1.5 \cdot {10}^{11} - {d}_{e b}}{\sqrt{\frac{1}{3} \cdot {10}^{6}}} = \frac{\left(1.5 \cdot {10}^{11} - {d}_{e b}\right) \cdot \sqrt{3}}{{10}^{3}}$

$\rightarrow {d}_{e b} \cdot {10}^{3} + \sqrt{3} \cdot {d}_{e b} = 1.5 \sqrt{3} \cdot {10}^{11}$

$\rightarrow {d}_{e b} \left({10}^{3} + \sqrt{3}\right) = \sqrt{6.75} \cdot {10}^{11}$

$\rightarrow {d}_{e b} = \frac{\sqrt{6.75} \cdot {10}^{11}}{{10}^{3} + \sqrt{3}} \approx 259358400$m

## How do astronomers use the Doppler effect to determine the velocities of astronomical objects?

Zack M.
8.736986301369862 years ago

Astronomers analyze the shift of spectral patterns of the light emitted or absorbed by those objects.

#### Explanation:

One of the problems which prompted Einstein's work on relativity was the constant speed of light in a vacuum. Classical physics would expect that even if the emission speed of light, $c$, were a constant, the observed speed would change with the relative velocity, $v$, of the light emitting object.

Laboratory observations, however, consistently measured the speed of light to be $3 \cdot {10}^{8} \text{ m/s}$. It turns out that the speed remains the same, but the wavelength is compressed or stretched depending on whether the object is moving toward or away from the observer.

Since the wavelength of light determines its color, we call this change "blueshift" for objects moving toward the observer, and "redshift" for objects moving away. Edwin Hubble derived a formula for measuring velocity based on the change in wavelength.

$v = \frac{\lambda - {\lambda}_{o}}{\lambda} _ o \cdot c$

This means that we need to know the emitted wavelength of the light in order to calculate the velocity. This is where spectroscopy comes in.

Every element on the periodic table has its own unique emission spectrum. This spectrum is formed when electrons within the atoms of these elements are excited to higher energies and then relax back to the ground state. In order to relax the atoms give off light. Because of quantum mechanics, however, electrons can only exist in specific orbital energies, so the atoms can only emit photons with wavelengths that correspond to the energies of these transitions.

Hydrogen is a convenient element to use for spectroscopy because not only does it have a fairly simple emission spectrum, it is abundant throughout the universe. If we analyze the light spectrum of a galaxy, we would expect to find an emission line at $656 \text{ nm}$. If instead we find that line at $670 \text{ nm}$, then the star is moving relative to us at a velocity of;

$v = \left(670 \cdot {10}^{-} 9 \text{ m" - 656* 10^-9 " m")/(656 * 10^-9 " m")(3*10^8 " m/s}\right)$

$= 6.4 \cdot {10}^{6} \text{ m/s}$

That's $6.4 \text{ million m/s}$ away from us. A similar process can be used to determine the speed of stars relative to us, but it involves absorption spectrums. The absorption spectrum of an element is the same as the emission spectrum except that instead of photons being emitted, they are subtracted from a background light source.

## A ball is thrown towards a cliff with an initial speed of 30.0 m/s directed at an angle of 60.0◦ above the horizontal. The ball lands on the edge of the cliff 4.00 s after it is thrown. ?

Michael
8.673972602739726 years ago

(a)

$25.5 \text{m}$

(b)

$34.4 \text{m}$

(c)

$20.0 \text{m/s}$ at an angle to the vertical of ${48.6}^{\circ}$

#### Explanation:

(a)

Diagram (a) describes the scenario. (apologies for the artwork):

(a)

To get $h$ we can use the equation of motion:

$s = u t + \frac{1}{2} a {t}^{2}$

The vertical component of the initial velocity $v$ is given by:

$u = v \sin 60$

So the expression for $h$ becomes:

$h = v \sin 60 t - \frac{1}{2} \text{g} {t}^{2}$

$\therefore h = 30 \sin 60 \times 4 - 0.5 \times 9.8 \times {4}^{2}$

$\therefore h = 103.9 - 78.4 = 25.5 \text{m}$

(b)

To get the maximum height ${h}_{\max}$ we can use:

${v}^{2} = {u}^{2} + 2 a s$

This becomes:

$0 = {\left(v \sin 60\right)}^{2} - 2 g {h}_{\max}$

$: {h}_{\max} = {\left(30 \times 0.866\right)}^{2} / \left(2 \times 9.8\right)$

$\therefore h = 34.43 \text{m}$

(c)

To get the vertical component ${v}_{y}$ of the impact velocity we need to get the time it takes to go from the maximum height to the moment it hits the cliff.

In diagram (a) the distance travelled is marked "y".

This can be found from:

$y = {h}_{\max} - h$

$\therefore y = 34.43 - 25.5 = 8.9 \text{m}$

So now we can say that;

$y = \frac{1}{2} \text{g} {t}^{2}$

:.t=sqrt((2y)/(g)

$\therefore t = \sqrt{\frac{2 \times 8.9}{9.8}} = 1.35 \text{s}$

Now we can use:

$v = u + a t$

This becomes:

${v}_{y} = 0 + \left(9.8 \times 1.35\right)$

$\therefore {v}_{y} = 13.2 \text{m/s}$

Now we know the vertical and horizontal components of velocity we can find the resultant ${v}_{r}$ by referring to diagram (b):

Using Pythagoras we can say that:

${v}_{r}^{2} = {13.2}^{2} + {\left(v \cos 60\right)}^{2}$

$\therefore {v}_{r}^{2} = {13.2}^{2} + {\left(30 \times 0.5\right)}^{2}$

${v}_{r}^{2} = 399.24$

$\therefore v = 20 \text{m/s}$

If you want the angle you can say that:

$\tan \alpha = \frac{v \cos 60}{13.2} = \frac{30 \times 0.5}{13.2} = 1.136$

From which $\alpha = {48.6}^{\circ}$

## Question 487bc

Truong-Son N.
8.638356164383561 years ago

Disclaimer: Yes, this will be long! No getting around it!

It really helps to know the derivations. So, what do we know?

• $\setminus m a t h b f \left(\mathrm{dH} = \mathrm{dU} + d \left(P V\right)\right)$ (1)
• $\setminus m a t h b f \left(\mathrm{dU} = \partial {q}_{\text{rev" + delw_"rev}}\right)$ (2)
• $\setminus m a t h b f \left(\partial {w}_{\text{rev}} = - P \mathrm{dV}\right)$ (3)
• \mathbf(((delH)/(delT))_P = C_P (4)
• \mathbf(((delU)/(delT))_V = C_V (5)

• An isothermal situation assumes a constant temperature during the expansion process.

• An adiabatic situation assumes no heat flow $\setminus m a t h b f \left(q\right)$ contributes to the internal energy $U$.
• The volume might have changed, but we don't know how, exactly. Even though we were given ${C}_{V}$, we can't assume that it is a constant-volume situation since we can convert from ${C}_{V}$ to ${C}_{p}$ pretty easily (${C}_{p} - {C}_{V} = n R$ for an ideal gas).

• The pressure decreased, i.e. it is NOT constant. That means that we should expect to eventually somehow use the relationship $P V = n R T$.

So, having said that, let's see...

---PART A---

ENTHALPY AND INTERNAL ENERGY

In an isothermal process, we know that $\Delta T = 0$. Using (4), we get:

$\mathrm{dH} = {C}_{p} \mathrm{dT}$

$\int \mathrm{dH} = \textcolor{b l u e}{\Delta H} = {\int}_{{T}_{1}}^{{T}_{2}} {C}_{p} \mathrm{dT} = \textcolor{b l u e}{0}$

...and using (5), we get:

$\mathrm{dU} = {C}_{V} \mathrm{dT}$

$\int \mathrm{dU} = \textcolor{b l u e}{\Delta U} = {\int}_{{T}_{1}}^{{T}_{2}} {C}_{V} \mathrm{dT} = \textcolor{b l u e}{0}$

...for an ideal gas. REMEMBER THIS:

"The energy of an ideal gas depends upon only the temperature" (McQuarrie, Ch. 19-4). In other words, when the temperature is constant, enthalpy and internal energy are both $0$ for an ideal gas.

REVERSIBLE HEAT FLOW

Now, solving for the reversible heat flow, using (1), that means $\Delta U = - \Delta \left(P V\right)$, so, using (2) with (1):

$\partial {q}_{\text{rev" + delw_"rev}} = - \left(P \mathrm{dV} + V \mathrm{dP}\right)$

and then using (3):

$\partial {q}_{\text{rev}} - \cancel{P \mathrm{dV}} = - \cancel{P \mathrm{dV}} - V \mathrm{dP}$

$\partial {q}_{\text{rev}} = - V \mathrm{dP}$

$\int \partial {q}_{\text{rev}} = - {\int}_{{P}_{1}}^{{P}_{2}} V \mathrm{dP}$

We don't know how the volume changed, just how the pressure changed, so we have to substitute $V$ for something else using the ideal gas law. So, $V = \frac{n R T}{P}$, where $n$, $R$, and $T$ are all constant:

$\textcolor{b l u e}{{q}_{\text{rev}}} = - {\int}_{{P}_{1}}^{{P}_{2}} \frac{n R T}{P} \mathrm{dP}$

$= - n R T {\int}_{{P}_{1}}^{{P}_{2}} \frac{1}{P} \mathrm{dP}$

$\textcolor{b l u e}{= - n R T \ln | \frac{{P}_{2}}{{P}_{1}} |}$

REVERSIBLE WORK

For reversible work, note that:

$\mathrm{dU} = \partial {q}_{\text{rev" + delw_"rev}}$

${\cancel{\Delta U}}^{0} = {q}_{\text{rev" + w_"rev}}$

thus:

$\textcolor{b l u e}{{w}_{\text{rev" = -q_"rev}}}$

---PART B---

REVERSIBLE HEAT FLOW, AND INTERNAL ENERGY

In an adiabatic process, we should know that $\textcolor{b l u e}{\partial {q}_{\text{rev}} = 0}$. Thus:

$\mathrm{dU} = {C}_{V} \mathrm{dT} = \partial {w}_{\text{rev}} = - P \mathrm{dV}$

In this case, ${T}_{1} = \text{300 K}$ and ${T}_{2} = \text{102 K}$. Furthermore, we again need to realize that the internal energy of an ideal gas depends only upon the temperature. Not the pressure, nor the volume. Therefore, we can use the definition $\mathrm{dU} = {C}_{V} \mathrm{dT}$:

$\textcolor{b l u e}{\Delta U} = {\int}_{{T}_{1}}^{{T}_{2}} {C}_{V} \mathrm{dT}$

$= \textcolor{b l u e}{\frac{3}{2} R \left({T}_{2} - {T}_{1}\right)}$

REVERSIBLE WORK

Next, we can use the relationship recently established to determine ${w}_{\text{rev}}$. Since $\partial {q}_{\text{rev}} = 0$, $\partial w = \mathrm{dw} = \mathrm{dU}$ (work becomes an exact differential), and:

$\textcolor{b l u e}{{w}_{\text{rev}} = \Delta U}$

ENTHALPY

Finally, we still need $\Delta H$! Note that again, the enthalpy depends only upon the temperature. Not the pressure, nor the volume. Thus, we can use the definition $\mathrm{dH} = {C}_{p} \mathrm{dT}$:

$\Delta H = {\int}_{{T}_{1}}^{{T}_{2}} {C}_{p} \mathrm{dT}$

Thus:

$\textcolor{b l u e}{\Delta H} = {C}_{p} \left({T}_{2} - {T}_{1}\right)$

$= \left({C}_{V} + n R\right) \left({T}_{2} - {T}_{1}\right)$

$= \left(\frac{3}{2} R + R\right) \left({T}_{2} - {T}_{1}\right)$

$= \textcolor{b l u e}{\frac{5}{2} R \left({T}_{2} - {T}_{1}\right)}$

## One "mol" of an ideal gas, initially at "300 K", is cooled at constant "V" so that "P"_f is 1/4 "P"_i. Then the gas expands at constant "P" until it reaches "T"_i again. What is the work w done on the gas?

Truong-Son N.
8.63013698630137 years ago

DISCLAIMER: Yes, this will be long!

This is one of those problems where drawing a "P-V diagram" would really help. Let's see approximately what that looks like for this problem.

We know that:

STEP 1
${T}_{i} = \text{300 K}$

${T}_{i} \downarrow \implies {T}_{f}$ such that:
${P}_{f} \downarrow \implies {P}_{i} \text{/} 4$
$\Delta V = 0$

STEP 2
${T}_{i} = \text{previous } {T}_{f}$
${V}_{i} = \text{previous } {V}_{i}$

${V}_{i} \uparrow \implies {V}_{f}$ such that:
${T}_{f} \uparrow \implies {T}_{i} = \text{300 K}$
$\Delta P = 0$

From that, we construct the PV-diagram as follows:

Notice how we have the possibility for a reversible work process that accomplishes the same thing more efficiently.

TWO-STEP PATH: STEP 1

We should know that inexact differential work $\partial w = - P \mathrm{dV}$ when expanding or compressing a gas, and that the inexact differential heat flow $\partial q$ is not $0$.

STEP 1: WORK

We can then proceed by noting that $\Delta V = 0$. Thus, any expansion/compression work is $0$:

$\textcolor{b l u e}{{w}_{1}} = - \int P \mathrm{dV} = \textcolor{b l u e}{0}$

STEP 1: HEAT FLOW

Since PV-work didn't happen here, it was all heat flow, conducting out through the surfaces of the container, slowly leading the gas particles to redistribute more loosely without changing in volume, thus naturally cooling the gas.

In other words, the change in internal energy was all due to heat flow at a constant volume ${q}_{V}$:

color(blue)(DeltaU = q_V = ???)

Unfortunately we don't know enough to determine ${q}_{V}$. Had we known whether the gas was monatomic, diatomic linear, or what have you, we could determine its degrees of freedom and write ${C}_{V}$ or ${C}_{p}$ in terms of the universal gas constant $R$, since basically:

\mathbf(DeltaU = q_V

$\setminus m a t h b f \left(= {\int}_{{T}_{1}}^{{T}_{2}} {C}_{v} \mathrm{dT}\right)$

Either way, this release of thermal energy means that the first work step was inefficient; thermal energy escaped the container while the work ${w}_{1}$ was being done.

TWO-STEP PATH: STEP 2

Okay, so what's ${w}_{2}$ then?

STEP 2: WORK

${w}_{2} = - \int P \mathrm{dV}$

$P$ is constant here, but the volume DOES change, so:

$\textcolor{g r e e n}{{w}_{2}} = - P {\int}_{{V}_{i}}^{{V}_{f}} \mathrm{dV}$

$= - P \left({V}_{f} - {V}_{i}\right)$

We don't know how the volume changed, but using the ideal gas law, we can fidget with this:

$= - \cancel{P} \left(\frac{n R {T}_{f}}{\cancel{P}} - \frac{n R {T}_{i}}{\cancel{P}}\right)$

$= \textcolor{g r e e n}{- n R \left({T}_{f} - {T}_{i}\right)}$

STEP 2: TEMPERATURE

Alright, so we know ${T}_{f} = \text{300 K}$ in step 2. What's ${T}_{i}$ in step 2? Well, we can use what we knew from step 1. With a constant volume in step 1:

$P V = n R T$

${P}_{i} {\cancel{{V}_{i}}}^{\text{constant") = cancel(nR)^("constant}} {T}_{i}$

${P}_{f} {\cancel{{V}_{i}}}^{\text{constant") = cancel(nR)^("constant}} {T}_{f}$

${P}_{i} / {T}_{i} = {P}_{f} / {T}_{f}$

The temperature from step 1 we want is ${T}_{f}$, which is ${T}_{i}$ in step 2, so:

${T}_{f} = \frac{{P}_{f}}{{P}_{i}} \times {T}_{i} = \frac{1}{4} \times 300 = \textcolor{g r e e n}{\text{75 K}}$

Thus, in step 2, the gas heated from ${T}_{i} = \text{75 K}$ to ${T}_{f} = \text{300 K}$. Finally, we're getting a value! So, with $R = \text{8.314472 J/mol"cdot"K}$:

$\textcolor{b l u e}{{w}_{2} = - \text{1870.76 J}}$

In expansion work, work is done by the gas, so it should be negative.

REVERSIBLE WORK PATH

Now, if we try that reversible path, we assume that steps 1 and 2 aren't there and that this is an efficient path where the gas is acted upon very slowly so that it has time to adjust. We know that in this path, $\Delta T = 0$, but $P$ and $V$ change. So:

REVERSIBLE WORK PATH: WORK

${w}_{\text{rev}} = - \int P \mathrm{dV}$

$= - \int \frac{n R T}{V} \mathrm{dV}$

$T$ is actually constant this time, so...

$\textcolor{g r e e n}{{w}_{\text{rev}}} = - n R T {\int}_{{V}_{i}}^{{V}_{f}} \frac{1}{V} \mathrm{dV}$

$= \textcolor{g r e e n}{- n R T \ln | \frac{{V}_{f}}{{V}_{i}} |}$

Notice how the negative natural logarithm approximates the shape of the reversible path.

REVERSIBLE WORK PATH: VOLUME

Okay, but how did the volume change? Inversely proportionally to the pressure! The volume increased at a constant temperature, so the pressure should decrease.

Using the same trick with the ideal gas law as before:

$\textcolor{g r e e n}{{w}_{\text{rev}}} = - n R T \ln | \frac{\frac{\cancel{n R T}}{P} _ f}{\frac{\cancel{n R T}}{P} _ i} |$

$= - n R T \ln | \frac{{P}_{i}}{{P}_{f}} |$

$= - n R T \ln | \frac{{P}_{i}}{\frac{1}{4} {P}_{i}} |$

$= \textcolor{g r e e n}{- n R T \ln 4}$

So, with $R = \text{8.314472 J/mol"cdot"K}$:

$\textcolor{b l u e}{{w}_{\text{rev" = -"3457.89 J}}}$

In expansion work, work is done by the gas, so it should be negative.

## A ball with a mass of  5 kg is rolling at 7 m/s and elastically collides with a resting ball with a mass of 4 kg. What are the post-collision velocities of the balls?

Gió
8.610958904109589 years ago

I found: $0.78 \mathmr{and} 7.78 \frac{m}{s}$ but check my maths!

#### Explanation:

We can use conservation of momentum $\vec{p} = m \vec{v}$ in one dimension (along $x$). So we get:

${p}_{\text{before")=p_("after}}$

So:

$\left(5 \cdot 7\right) + \left(4 \cdot 0\right) = \left(5 \cdot {v}_{1}\right) + \left(4 \cdot {v}_{2}\right)$

Being an elastic collision also kinetic energy $K = \frac{1}{2} m {v}^{2}$ is conserved:

${K}_{\text{before")=K_("after}}$

So:
$\frac{1}{2} \cdot 5 \cdot {7}^{2} + \frac{1}{2} \cdot 4 \cdot {0}^{2} = \frac{1}{2} \cdot 5 \cdot {v}_{1}^{2} + \frac{1}{2} \cdot 4 \cdot {v}_{2}^{2}$

We can use the two equations together and we get:

$\left\{\begin{matrix}35 = 5 {v}_{1} + 4 {v}_{2} \\ 245 = 5 {v}_{1}^{2} + 4 {v}_{2}^{2}\end{matrix}\right.$

From the first equation:
${v}_{1} = \frac{35 - 4 {v}_{2}}{5}$
Substitute into the second and get:
$245 = \cancel{5} {\left(35 - 4 {v}_{2}\right)}^{2} / {\cancel{25}}^{5} + 4 {v}_{2}^{2}$
$\cancel{1225} = \cancel{1225} - 280 {v}_{2} + 16 {v}_{2}^{2} + 20 {v}_{2}^{2}$
$36 {v}_{2}^{2} - 280 {v}_{2} = 0$
${v}_{2} = 0 \frac{m}{s}$ (not) and ${v}_{2} = \frac{280}{36} = \frac{70}{9} = 7.78 \frac{m}{s}$ yes.
Corresponding to:
${v}_{1} = 7 \frac{m}{s}$ and ${v}_{1} = \frac{7}{9} = 0.78 \frac{m}{s}$

## A charge of 7 C is at the origin. How much energy would be applied to or released from a  5 C charge if it is moved from  (-3 , 7 )  to (-7 ,-2 ) ?

Michael
8.608219178082193 years ago

$0.565 \text{k}$ units

#### Explanation:

I'll work out the electric potential at points P and Q.

Then I will use this to work out the potential difference between the 2 points.

This is the work done by moving a unit charge between the 2 points.

The work done in moving a 5C charge between P and Q can therefore be found by multiplying the potential difference by 5.

We need to find the length of PR and PQ which can be done using Pythagoras:

$P {R}^{2} = {2}^{2} + {7}^{2}$

$\therefore P R = \sqrt{4 + 49}$

$P R = 7.8$

and:

$Q {R}^{2} = {3}^{2} + {7}^{2}$

$\therefore Q R = \sqrt{9 + 49} = 6.93$

The electric potential due to a charge $q$ at a distance $r$ is given by:

$V = k . \frac{q}{r}$

$k$ is a constant.

So the potential at point $P$ is given by:

${V}_{P} = k \times \frac{7}{7.8} = 0.897 k$

The potential at $Q$ is given by:

${V}_{Q} = k \times \frac{7}{6.93} = 1.01 k$

So the potential difference is given by:

${V}_{P} - {V}_{Q} = k \left(0.897 - 1.01\right) = - 0.113 k$

So the work done in moving a 5C charge between these 2 points is given by:

$W = - 0.113 k \times 5 = - 0.565 k \text{units}$

This is the work done on the charge.

There are no units of distance given. If this was in meters then $k = 9 \times {10}^{9} \text{m/F}$ and the answer would be in Joules.

## A projectile is shot from the ground at an angle of (5 pi)/12  and a speed of 3 m/s. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

Zack M.
8.605479452054794 years ago

$.230$ meters from the starting point.

#### Explanation:

We should start by breaking down the initial velocity into its $x$ and $y$ components. We can find the projection of $\vec{V}$ onto the $x$ and $y$ axis by using $\cos \theta$ and $\sin \theta$ respectively.

$\vec{V} = \vec{V} \cdot \hat{x} + V \cdot \hat{y}$

$= \left(V \cos \theta\right) \hat{x} + \left(V \sin \theta\right) \hat{y}$

$= 3 \cos \left(\frac{5 \pi}{12}\right) \hat{x} + 3 \sin \left(\frac{5 \pi}{12}\right) \hat{y}$

$= .776 \hat{x} + 2.898 \hat{y}$

Note that the units are $\text{m/s}$. The initial velocity is mostly in the $y$ direction, so we shouldn't expect the projectile to go very far. Now we can calculate how long it takes for the projectile to reach the highest point. At its highest, the vertical velocity of the projectile will be zero. We can use the following equation to find at what time the velocity is zero.

${V}_{y} \left(t\right) = a t + {V}_{\circ} = 0$

$t = - {V}_{\circ} / a$

The acceleration due to gravity is $- 9.8 \frac{m}{s} ^ 2$.

$t = - \frac{2.989}{- 9.8}$

$t = .296$

So the projectile takes $.296 \text{ seconds}$ to reach the top of its flight. Now we can plug this into the equation of motion for the $x$ direction. Since there is no net force on the projectile in the horizontal direction, the acceleration will be zero. Also, we are starting at the origin, so the initial $x$ position will be zero.

$x \left(t\right) = \frac{1}{2} {\cancel{a}}^{0} {t}^{2} + {V}_{x} t + {\cancel{{x}_{\circ}}}^{0}$

$x \left(.296\right) = \left(.776\right) \left(.296\right)$

$x \left(.296\right) = .230$

The projectile will move about $.230 \text{ m}$ from the starting point when it reaches the top of its flight.

## If a 13 kg object moving at 7 m/s slows down to a halt after moving 10 m#, what is the friction coefficient of the surface that the object was moving over?

Trevor Ryan.
8.602739726027398 years ago

${\mu}_{k} = 0.25$

#### Explanation:

I will show you 2 different methods to do this question :

1. Method 1 - Using Newton's Laws and Equations of Motion :

The acceleration of the object is uniform and in 1 direction and can be found from a relevant equation of motion for constant acceleration in 1 dimension as follows :

${v}^{2} = {u}^{2} + 2 a x$

$\therefore a = \frac{{v}^{2} - {u}^{2}}{2 x} = \frac{{0}^{2} - {7}^{2}}{2 \times 10} = - 2.45 m / {s}^{2}$.

This is acceleration is caused by the frictional force which is the resultant force acting on the object and so by Newton 2 and definition of frictional forces we get :

$\sum \vec{F} = m \vec{a}$

$\therefore - {f}_{k} = m a$, where ${f}_{k} = {\mu}_{k} N$.

But $N = m g \implies {\mu}_{k} m g = m a$.

$\therefore {\mu}_{k} = \frac{a}{g} = \frac{2.45}{9.8} = 0.25$.

2. Method 2 - Using energy considerations :

The Work-Energy Theorem states that the work done by the resultant force equals the change in kinetic energy brought about.

$\therefore {W}_{{f}_{k}} = \Delta {E}_{k}$

$\therefore {f}_{k} \cdot x = \frac{1}{2} m \left({v}^{2} - {u}^{2}\right)$

$\therefore {\mu}_{k} m g \cdot x = \frac{1}{2} m \left({v}^{2} - {u}^{2}\right)$

$\therefore {\mu}_{k} = \frac{{v}^{2} - {u}^{2}}{2 g x} = \frac{{0}^{2} - {7}^{2}}{2 \times 9.8 \times 10} = 0.25$.