7.104109589041096 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!!

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.

6.997260273972603 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.

The gravitational attraction between two bodies is calculated as follows:

with

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

We are considering the situation when

Knowing that

6.854794520547945 years ago

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

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,

Laboratory observations, however, consistently measured the speed of light to be

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.

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

That's

6.791780821917809 years ago

(a)

(b)

(c)

(a)

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

(a)

To get

The vertical component of the initial velocity

So the expression for

(b)

To get the maximum height

This becomes:

(c)

To get the vertical component

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

This can be found from:

So now we can say that;

Now we can use:

This becomes:

Now we know the vertical and horizontal components of velocity we can find the resultant

Using Pythagoras we can say that:

If you want the angle you can say that:

From which

6.756164383561644 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 **(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 anideal gasdepends upon"only the temperature(McQuarrie, Ch. 19-4).In other words, when the temperature is constant, enthalpy and internal energy areboth#0# for anideal gas.

**REVERSIBLE HEAT FLOW**

Now, solving for the reversible heat flow, using **(1)**, that means **(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 ** 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

#dU = C_VdT = delw_"rev" = -PdV#

In this case,

#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 **exact differential**), and:

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

**ENTHALPY**

Finally, we still need

#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))#

6.747945205479452 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**

**STEP 2**

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** **inexact differential heat flow**

**STEP 1: WORK**

We can then proceed by noting that

#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

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

Unfortunately we don't know enough to determine

#\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

**TWO-STEP PATH: STEP 2**

Okay, so what's

**STEP 2: WORK**

#w_2 = -int PdV#

#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 **2**. What's **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 **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

#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,

**REVERSIBLE WORK PATH: WORK**

#w_"rev" = -int PdV#

#= -int (nRT)/V dV#

#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

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

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

6.7315068493150685 years ago

I found:

We can use conservation of momentum

So:

Being an elastic collision also kinetic energy

So:

We can use the two equations together and we get:

From the first equation:

Substitute into the second and get:

Corresponding to:

6.728767123287671 years ago

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:

and:

The electric potential due to a charge

So the potential at point

The potential at

So the potential difference is given by:

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

This is the work done on the charge.

There are no units of distance given. If this was in meters then

6.723287671232876 years ago

We should start by breaking down the initial velocity into its

Note that the units are

The acceleration due to gravity is

So the projectile takes

The projectile will move about

6.723287671232876 years ago

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 :

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 :

But

**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.