VCE Physics Question Thread!

[Alexicology]

How do you do q8,9
Thanks

Hi, I'm a current Year 12 student,

For question 8, use Coulomb's Law to find the force of repulsion. So F=kq1q2/r^2

Sub in k= 9x10^9, q1=1.602x10^-19, q2=1.602x10^-19 and r=2.5x10^-15

Therefore, force of repulsion= 37N


For Q9, use F=k1k2/r2 again to find q1

q1= 6.24x10^7C

Then find number of electrons transfered by q1/qe

Therefore number of electrons= 3.9x10^26 electrons

[Mattjbr2]

Can someone please confirm for me the correct answers to these questions, as well as explain how they arrived at said answers?
I don't understand why the book says the answers for 3 and 4 aren't D and B, respectively.

Edit: The book claims C and C

[Syndicate]

Can someone please confirm for me the correct answers to these questions, as well as explain how they arrived at said answers?
I don't understand why the book says the answers for 3 and 4 aren't D and B, respectively.

Edit: The book claims C and C

You are correct.

[Mattjbr2]

You are correct.

You sure?

[Syndicate]

You sure?

Yep.

Since Ffricition = 0N, the horizontal component of the normal force must be responsible for the net force. The horizontal component of the normal force is directly in front of the angle (which is vertically opposite to the angle beneath).

4) N = mg/cos15
mg = Ncos15

[dylang99]

The Sun orbits the centre of our galaxy, the Milky Way, at a distance of 2.2 × 1020 m from the centre with a period of 2.5 × 108 years.  e mass of all the stars inside the Sun’s orbit can be considered as being concentrated at the centre of the galaxy.  e mass of the Sun is 2.0 × 1030 kg. If all the stars have the same mass as the Sun, how many stars are in the Milky Way?

Any help would be greatly appreciated!

[Syndicate]

The Sun orbits the centre of our galaxy, the Milky Way, at a distance of 2.2 × 1020 m from the centre with a period of 2.5 × 108 years.  e mass of all the stars inside the Sun’s orbit can be considered as being concentrated at the centre of the galaxy.  e mass of the Sun is 2.0 × 1030 kg. If all the stars have the same mass as the Sun, how many stars are in the Milky Way?

Any help would be greatly appreciated!

Number of stars = total mass of stars/ mass of sun (as each star has an equal weight).

I am going to assume that you mean the number of stars in a sphere (where the sphere = the total area between the sun's orbit and the milky way).
Msun = \(2.0 \times 10^{30} \) kg
T = \( 7.884 \times 10^{15} \) seconds (assuming each year has 365 days)
r = \( 2.2 \times 10^{20} \) metres
M = total mass of stars

So firstly we need to calculate the mass of the stars in the area by equating Fg to Fc because all of the masses of the stars are concentrated at the centre of the galaxy.

[Gogo14]

How do u do this q?

[cosecant]

How do u do this q?

since the electrons emerge with no deflection and are subjected to both the magnetic field and electric field, the magnetic force must be balanced by the electric force, thus qE=qvB. Hence, E=vB

 since E=V/d  (we are trying to find the distance - d)
-> V/d=vB. d is the unknown, we have V and B so we need to find v first.

since work is being done on the electrons by the electric field
qV= (mv²)/2
q =1.6*10^-19, m=9.1*10^-31, V = 3*10³
solving, v equals to 3.248*10⁷ m/s

sub v= 3.248*10⁷, V=3*10³ and B= 1.6*10^-3 into the equatio previously derived  V/d=vB
rearranging, d=V/vB
d= (3*10³)/(3.248*10⁷*1.6*10^-3)
d= 0.0577m
=5.8cm

[TooLazy]

Could someone please explain how to get the answer in Q14,
and why it is the answer

[Shadowxo]

Could someone please explain how to get the answer in Q16,
and why it is the answer

You only posted 14, do you want to know 14 or 16?

[TooLazy]

You only posted 14, do you want to know 14 or 16?

Sorry, I meant 14

[wyzard]

Could someone please explain how to get the answer in Q14,
and why it is the answer

Simply equate the gravitational forces acting on X by each planet using Newton's law of Gravity, then you'll find the term G and R cancels out, leaving you only M and m. Rearranging the M and m, you can find M/m to be 16.

[nicholas9027]

Hey guys
I need help on this Wave Question
1.Calculate the longest and shortest time for a radio signal travelling at the speed of light to go from the Earth to a space probe when the space probe is a) near Mars and (b) near Neptune.

Radius of Earth's orbit about the Sun= 1.49 x 10^11m
Radius of Mar's orbit about the Sun=2.28 x 10^11m
Radius of Neptune's orbit about the Sun= 4.50 x 10^12m

I've found the radius between Earth and Mars is 0.79 x 10^11m i think.
Thanks in advance 

[Shadowxo]

Hey guys
I need help on this Wave Question
1.Calculate the longest and shortest time for a radio signal travelling at the speed of light to go from the Earth to a space probe when the space probe is a) near Mars and (b) near Neptune.

Radius of Earth's orbit about the Sun= 1.49 x 10^11m
Radius of Mar's orbit about the Sun=2.28 x 10^11m
Radius of Neptune's orbit about the Sun= 4.50 x 10^12m

I've found the radius between Earth and Mars is 0.79 x 10^11m i think.
Thanks in advance 

I'm presuming this is the way they want you to do this.
a) So the shortest distance between Mars and Earth is when there is a straight line connecting the sun, earth, and mars, ie earth is partway between sun and mars, so they're on the same side of the sun and they're closest. So, the distance would be 2.28 x 10¹¹ - 1.49 x 10¹¹ = 7.9 x 10¹⁰ which is what you found out. Just use the speed of sound to figure out the time it would take to travel that distance.
The longest distance between Mars and Earth is when they're on opposite sides of the sun, directly opposite each other. So the distance between them would be 2.28 x 10¹¹ + 1.49 x 10¹¹ = 3.77 x 10¹¹ (distance between Mars and Sun + distance between Earth and Sun). Then, do the same thing to determine the time it would take to travel that distance.
Repeat for b)