HSC Physics Question Thread

[RuiAce]

Will just clarify that that last line should read:



Note the change in units and the extra 0 

Couldn't you just have edited my mistake (please)

[jamonwindeyer]

Just curious but why is "g" (or whatever value it was, I forgot) always 9.8 ms^-2?

Thanks

That is the average value of the acceleration due to gravity on earth (it does vary slightly with altitude and location). If we are on earth's surface, that is a physical constant, meaning it is the pretty much the same everywhere on earth 

[RuiAce]

Just curious but why is "g" (or whatever value it was, I forgot) always 9.8 ms^-2?

Thanks

It's 9.8 m s^-2 on EARTH.

(In actuality it isn't, because the Earth is not a sphere - rather a spheroid. We just assume the Earth is a sphere for simplicity.)

[jamonwindeyer]

Couldn't you just have edited my mistake (please)

With your permission I will definitely do that from now on (don't like to touch other peoples posts without permission - unless I have to of course - arbitrary use of moderator power  )

[RuiAce]

With your permission I will definitely do that from now on (don't like to touch other peoples posts without permission - unless I have to of course - arbitrary use of moderator power  )

Aha if it's just a correction then I permit just make a note of what you fixed please

[EEEEEEP]

Just curious but why is "g" (or whatever value it was, I forgot) always 9.8 ms^-2?

Thanks

BEcause that's how strong gravity is on earth.

[conic curve]

That is the average value of the acceleration due to gravity on earth (it does vary slightly with altitude and location). If we are on earth's surface, that is a physical constant, meaning it is the pretty much the same everywhere on earth 

So this is "by definition"

[EEEEEEP]

So this is "by definition"

NOt by definition. But it is the standard figure used for problems.

[conic curve]

NOt by definition. But it is the standard figure used for problems.

Where's it derived from? It must be "by convention" then

It doesn't really say here: http://www.physicsclassroom.com/class/1DKin/Lesson-5/Acceleration-of-Gravity

[RuiAce]

Where's it derived from? It must be "by convention" then

It doesn't really say here: http://www.physicsclassroom.com/class/1DKin/Lesson-5/Acceleration-of-Gravity

Jamon already answered this.

That is the average value of the acceleration due to gravity on earth (it does vary slightly with altitude and location). If we are on earth's surface, that is a physical constant, meaning it is the pretty much the same everywhere on earth 

[EEEEEEP]

Where's it derived from? It must be "by convention" then

It doesn't really say here: http://www.physicsclassroom.com/class/1DKin/Lesson-5/Acceleration-of-Gravity

It's not important to know how it is derived. They don't ask you to derive the gravity. Here it is anyway.

 Newtons law states that the sum of the forces equals the mass* acceleration


m*a = F

gravitational force is calculated as

F = G*m*M/r^2

where G is the gravitational constant (measured, look it up on wikipedia)
m is the mass of the falling body
M is the mass of the earth (or other planet)
r is the distance between the falling body and the center of the earth (or other planet)

m*a = G*m*M/r^2

cancel m

a = G*M/r^2

if you put in the correct values

G = 6.67428*10^(-11) m^3/(kgs^2)
M = 5.9736*10^(24) kg
r = 6.371*10^(6) m - average radius of earth

so if the mass is approximatly on the surface of the earth (say only a few feet above it) then the acceleration is

a = 9.8226 m/s^2

pretty close to the measured value of 9.80667---m/s^2. You'd get closer to that value if you used more significant digits in G, M and r.

[RuiAce]

It's not important to know how it is derived. They don't ask you to derive the gravity. Here it is anyway.

[jakesilove]

It's not important to know how it is derived. They don't ask you to derive the gravity. Here it is anyway.

 Newtons law states that the sum of the forces equals the mass* acceleration


m*a = F

gravitational force is calculated as

F = G*m*M/r^2

where G is the gravitational constant (measured, look it up on wikipedia)
m is the mass of the falling body
M is the mass of the earth (or other planet)
r is the distance between the falling body and the center of the earth (or other planet)

m*a = G*m*M/r^2

cancel m

a = G*M/r^2

if you put in the correct values

G = 6.67428*10^(-11) m^3/(kgs^2)
M = 5.9736*10^(24) kg
r = 6.371*10^(6) m - average radius of earth

so if the mass is approximatly on the surface of the earth (say only a few feet above it) then the acceleration is

a = 9.8226 m/s^2

pretty close to the measured value of 9.80667---m/s^2. You'd get closer to that value if you used more significant digits in G, M and r.

Was literally about to post the derivation, thanks for sending that through! It's not too difficult a proof, even if it isn't directly examinable. A bit of cool Physics

[MysteryMarker]

An object that is travelling at a speed of 0.9c relative to an observer has a mass of 100kg measured by the observer. Calculate the mass of the object measured by the same observer when it is at rest relative to the observer?

Why is the answer to this question not 100kg? I know the formula and how to get the answer, i just don't under stand why we still have to use the formula.

Thanks guys.

[RuiAce]

An object that is travelling at a speed of 0.9c relative to an observer has a mass of 100kg measured by the observer. Calculate the mass of the object measured by the same observer when it is at rest relative to the observer?

Why is the answer to this question not 100kg? I know the formula and how to get the answer, i just don't under stand why we still have to use the formula.

Thanks guys.

Can't explain it carefully anymore but the idea is to keep in mind that all frames of references are relative. Mass dilation says that moving objects become heavier. If the object is moving at 0.9c, it is going to appear way heavier than if it were at rest (relatively speaking).