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Question #bfcde

Zack M.
Zack M.
9.512328767123288 years ago

Answer:

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.

enter image source here

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.

enter image source here

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.

enter image source here

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.

enter image source here

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.
Olivier B.
9.402739726027397 years ago

Answer:

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=(G*m_1*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_(sb)=(G*m_s*m_b)/((d_(sb))^2)

F_(eb)=(G*m_e*m_b)/((d_(eb))^2)

We are considering the situation when F_(sb)=F_(eb):

(cancel(G)*m_s*cancel(m_b))/((d_(sb))^2)=(cancel(G)*m_e*cancel(m_b))/((d_(eb))^2)

rarr (m_s)/((d_(sb))^2)=(m_e)/((d_(eb))^2)

rarr (m_s)/(m_e)=((d_(sb))^2)/((d_(eb))^2)=((d_(sb))/(d_(eb)))^2

rarr sqrt((m_s)/(m_e))=(d_(sb))/(d_(eb))

Knowing that m_e=6*10^24kg and m_s=2*10^30kg, and that d_(sb)+d_(eb)=1.5*10^11m we have to solve the following equation:

sqrt((2*10^30)/(6*10^24))=(1.5*10^11-d_(eb))/(d_(eb))

rarr d_(eb)=(1.5*10^11-d_(eb))/(sqrt((2*10^30)/(6*10^24)))=(1.5*10^11-d_(eb))/(sqrt(1/3*10^6))=((1.5*10^11-d_(eb))*sqrt3)/(10^3)

rarr d_(eb)*10^3+sqrt(3)*d_(eb)=1.5sqrt3*10^11

rarr d_(eb)(10^3+sqrt3)=sqrt6.75*10^11

rarr d_(eb)=(sqrt6.75*10^11)/(10^3+sqrt3)~~259358400m

Answer:

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.

enter image source here

Laboratory observations, however, consistently measured the speed of light to be 3*10^8 " 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.

enter image source here

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 = (lambda - lambda_o)/lambda_o* 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.

http://ch301.cm.utexas.edu/section2.php?target=atomic/H-atom/line-spectra.html

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 " nm". If instead we find that line at 670 " nm", then the star is moving relative to us at a velocity of;

v = (670 * 10^-9" m" - 656* 10^-9 " m")/(656 * 10^-9 " m")(3*10^8 " m/s")

= 6.4*10^6 " m/s"

That's 6.4 " 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.

http://ch301.cm.utexas.edu/section2.php?target=atomic/H-atom/line-spectra.html

Answer:

(a)

25.5"m"

(b)

34.4"m"

(c)

20.0"m/s" at an angle to the vertical of 48.6^@

Explanation:

(a)

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

MFdocs

(a)

To get h we can use the equation of motion:

s=ut+1/2at^2

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

u=vsin60

So the expression for h becomes:

h=vsin60t-1/2"g"t^2

:.h=30sin60xx4-0.5xx9.8xx4^2

:.h=103.9-78.4=25.5"m"

(b)

To get the maximum height h_(max) we can use:

v^2=u^2+2as

This becomes:

0=(vsin60)^2-2gh_(max)

:h_(max)=(30xx0.866)^2/(2xx9.8)

:.h=34.43"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

:.y=34.43-25.5=8.9"m"

So now we can say that;

y=1/2"g"t^2

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

:.t=sqrt((2xx8.9)/(9.8))=1.35"s"

Now we can use:

v=u+at

This becomes:

v_(y)=0+(9.8xx1.35)

:.v_(y)=13.2"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+(vcos60)^2

:.v_r^2=13.2^2+(30xx0.5)^2

v_r^2=399.24

:.v=20"m/s"

If you want the angle you can say that:

tanalpha=(vcos60)/13.2=(30xx0.5)/13.2=1.136

From which alpha=48.6^@

Question #487bc

Truong-Son N.
Truong-Son N.
9.161643835616438 years ago

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


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

  • \mathbf(dH = dU + d(PV)) (1)
  • \mathbf(dU = delq_"rev" + delw_"rev") (2)
  • \mathbf(delw_"rev" = -PdV) (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 \mathbf(q) 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 = nR 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 PV = nRT.

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

---PART A---

ENTHALPY AND INTERNAL ENERGY

In an isothermal process, we know that DeltaT = 0. Using (4), we get:

dH = C_pdT

int dH = color(blue)(DeltaH) = int_(T_1)^(T_2) C_pdT = color(blue)(0)

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

dU = C_VdT

int dU = color(blue)(DeltaU) = int_(T_1)^(T_2) C_VdT = color(blue)(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 DeltaU = -Delta(PV), so, using (2) with (1):

delq_"rev" + delw_"rev" = -(PdV + VdP)

and then using (3):

delq_"rev" - cancel(PdV) = - cancel(PdV) - VdP

delq_"rev" = -VdP

intdelq_"rev" = -int_(P_1)^(P_2) VdP

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 = (nRT)/P, where n, R, and T are all constant:

color(blue)(q_"rev") = -int_(P_1)^(P_2) (nRT)/P dP

= -nRT int_(P_1)^(P_2) 1/P dP

color(blue)(= -nRT ln|(P_2)/(P_1)|)

REVERSIBLE WORK

For reversible work, note that:

dU = delq_"rev" + delw_"rev"

cancel(DeltaU)^(0) = q_"rev" + w_"rev"

thus:

color(blue)(w_"rev" = -q_"rev")

---PART B---

REVERSIBLE HEAT FLOW, AND INTERNAL ENERGY

In an adiabatic process, we should know that color(blue)(delq_"rev" = 0). Thus:

dU = C_VdT = delw_"rev" = -PdV

In this case, T_1 = "300 K" and T_2 = "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 dU = C_VdT:

color(blue)(DeltaU) = int_(T_1)^(T_2) C_VdT

= color(blue)(3/2 R (T_2 - T_1))

REVERSIBLE WORK

Next, we can use the relationship recently established to determine w_"rev". Since delq_"rev" = 0, delw = dw = dU (work becomes an exact differential), and:

color(blue)(w_"rev" = DeltaU)

ENTHALPY

Finally, we still need DeltaH! Note that again, the enthalpy depends only upon the temperature. Not the pressure, nor the volume. Thus, we can use the definition dH = C_pdT:

DeltaH = int_(T_1)^(T_2) C_pdT

Thus:

color(blue)(DeltaH) = C_p(T_2 - T_1)

= (C_V + nR)(T_2 - T_1)

= (3/2 R + R)(T_2 - T_1)

= color(blue)(5/2 R(T_2 - T_1))

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 = "300 K"

T_i darr => T_f such that:
P_f darr => P_i"/"4
DeltaV = 0

STEP 2
T_i = "previous " T_f
V_i = "previous " V_i

V_i uarr => V_f such that:
T_f uarr => T_i = "300 K"
DeltaP = 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 delw = -PdV when expanding or compressing a gas, and that the inexact differential heat flow delq is not 0.

STEP 1: WORK

We can then proceed by noting that DeltaV = 0. Thus, any expansion/compression work is 0:

color(blue)(w_1) = - int PdV = color(blue)(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

\mathbf(= int_(T_1)^(T_2) C_v dT)

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 PdV

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

color(green)(w_2) = -P int_(V_i)^(V_f) dV

= -P(V_f - V_i)

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

= -cancel(P)((nRT_f)/cancel(P) - (nRT_i)/cancel(P))

= color(green)(-nR(T_f - T_i))

STEP 2: TEMPERATURE

Alright, so we know T_f = "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:

PV = nRT

P_icancel(V_i)^("constant") = cancel(nR)^("constant")T_i

P_fcancel(V_i)^("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 = (P_f)/(P_i)xxT_i = 1/4 xx 300 = color(green)("75 K")

Thus, in step 2, the gas heated from T_i = "75 K" to T_f = "300 K". Finally, we're getting a value! So, with R = "8.314472 J/mol"cdot"K":

color(blue)(w_2 = -"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, DeltaT = 0, but P and V change. So:

REVERSIBLE WORK PATH: WORK

w_"rev" = -int PdV

= -int (nRT)/V dV

T is actually constant this time, so...

color(green)(w_"rev") = -nRT int_(V_i)^(V_f) 1/V dV

= color(green)(-nRTln|(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:

color(green)(w_"rev") = -nRTln|(cancel(nRT)/P_f)/(cancel(nRT)/P_i)|

= -nRTln|(P_i)/(P_f)|

= -nRTln|(P_i)/(1/4 P_i)|

= color(green)(-nRTln4)

So, with R = "8.314472 J/mol"cdot"K":

color(blue)(w_"rev" = -"3457.89 J")

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

Answer:

I found: 0.78 and 7.78 m/s but check my maths!

Explanation:

We can use conservation of momentum vecp=mvecv in one dimension (along x). So we get:

p_("before")=p_("after")

So:

(5*7)+(4*0)=(5*v_1)+(4*v_2)

Being an elastic collision also kinetic energy K=1/2mv^2 is conserved:

K_("before")=K_("after")

So:
1/2*5*7^2+1/2*4*0^2=1/2*5*v_1^2+1/2*4*v_2^2

We can use the two equations together and we get:

{(35=5v_1+4v_2),(245=5v_1^2+4v_2^2):}

From the first equation:
v_1=(35-4v_2)/5
Substitute into the second and get:
245=cancel(5)(35-4v_2)^2/cancel(25)^5+4v_2^2
cancel(1225)=cancel(1225)-280v_2+16v_2^2+20v_2^2
36v_2^2-280v_2=0
v_2=0m/s (not) and v_2=280/36=70/9=7.78m/s yes.
Corresponding to:
v_1=7m/s and v_1=7/9=0.78m/s

Answer:

0.565"k" units

Explanation:

MFDocs

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:

PR^2=2^2+7^2

:.PR=sqrt(4+49)

PR=7.8

and:

QR^2=3^2+7^2

:.QR=sqrt(9+49)=6.93

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

V=k.q/r

k is a constant.

So the potential at point P is given by:

V_P=kxx7/7.8=0.897k

The potential at Q is given by:

V_Q=kxx7/6.93=1.01k

So the potential difference is given by:

V_P-V_Q=k(0.897-1.01)=-0.113k

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

W=-0.113kxx5=-0.565k"units"

This is the work done on the charge.

There are no units of distance given. If this was in meters then k=9xx10^9"m/F" and the answer would be in Joules.

Answer:

.230 meters from the starting point.

Explanation:

enter image source here

We should start by breaking down the initial velocity into its x and y components. We can find the projection of vecV onto the x and y axis by using cos theta and sin theta respectively.

vecV =vecV *hat x + V*hat y

=(V cos theta) hat x + (V sin theta) hat y

= 3cos((5pi)/12) hatx + 3sin((5pi)/12) haty

= .776 hat x + 2.898 hat y

Note that the units are "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(t) = at + V_@ = 0

t=-V_@/a

The acceleration due to gravity is -9.8 m/s^2.

t = -(2.989)/(-9.8)

t = .296

So the projectile takes .296 " 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(t) = 1/2 cancel(a)^0 t^2 + V_x t + cancel(x_@)^0

x(.296) = (.776)(.296)

x(.296) = .230

The projectile will move about .230 " m" from the starting point when it reaches the top of its flight.

Answer:

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+2ax

therefore a=(v^2-u^2)/(2x)=(0^2-7^2)/(2xx10)=-2.45m//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 :

sumvecF=mveca

therefore -f_k=ma, where f_k=mu_kN.

But N=mg =>mu_kmg=ma.

therefore mu_k=a/g=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)=DeltaE_k

therefore f_k*x=1/2m(v^2-u^2)

therefore mu_kmg*x=1/2m(v^2-u^2)

therefore mu_k=(v^2-u^2)/(2gx)=(0^2-7^2)/(2xx9.8xx10)=0.25.