VCE Physics Question Thread!

[Syndicate]

To calculate force do I use F=(mv^2)/r
where m is 5g, v is 500m/s and the r is 50cm?

the answer is apparently around twenty but i cant seem to get it using this

(0.005x500^2)/0.5 = 2500

2500 = 100 x 4pi(0.45^2) x gram of gunpowder used?

the answer wouldve been 9.8 if that was the case... what am i doing wrong?

You are not trying to find the centripetal force (there is none in this case)

As Iznxl said, you need to use the formula 100/aw

To workout a, you must use the formula (believing it's a sphere): pi x r^2 (1.41 cm)

[lzxnl]

well offcourse... xD (unless it's fully scratched  )

yea, but he only asked if why the train tracks are banked (when going through a curve)
It's good to learn something new everyday 

My point was that you didn't connect banked curves with centripetal force. A non-banked curve has to wholly rely on friction for the centripetal force, whereas with a banked curve, you don't need as much friction.

To calculate force do I use F=(mv^2)/r
where m is 5g, v is 500m/s and the r is 50cm?

the answer is apparently around twenty but i cant seem to get it using this

(0.005x500^2)/0.5 = 2500

2500 = 100 x 4pi(0.45^2) x gram of gunpowder used?

the answer wouldve been 9.8 if that was the case... what am i doing wrong?

Think about what information the question is giving you. You have constant acceleration (don't know magnitude), known initial and final velocities and a known distance. That's enough to give you the value of the acceleration. As you know the force is some number divided by a known area and unknown mass of gunpowder and you know the force, you can work out the gunpowder mass.

[knightrider]

For this question, assume that:
mass of Earth = 6.0 × 10^24 kg
radius of Earth = 6.4 × 10^6 m
mass of the Moon = 7.3 × 10^22 kg
mean radius of Moon’s orbit = 3.8 × 10^8 m
Calculate the gravitational force of attraction that
exists between the following objects.

d ) The Moon and the Earth

solution is as follows

Gravitational force F
= (6.67 x 10^-11 × 7.3 x 10^22 × 6.0 x 10^24 )/(3.8 x 10^8)^2
= 2.0 x 10^20 N

shouldnt R be 6.4x10^6+3.8x10^8=3.9x10^8 m instead of 3.8 x 10^8  ?

[lzxnl]

For this question, assume that:
mass of Earth = 6.0 × 10^24 kg
radius of Earth = 6.4 × 10^6 m
mass of the Moon = 7.3 × 10^22 kg
mean radius of Moon’s orbit = 3.8 × 10^8 m
Calculate the gravitational force of attraction that
exists between the following objects.

d ) The Moon and the Earth

solution is as follows

Gravitational force F
= (6.67 x 10^-11 × 7.3 x 10^22 × 6.0 x 10^24 )/(3.8 x 10^8)^2
= 2.0 x 10^20 N

shouldnt R be 6.4x10^6+3.8x10^8=3.9x10^8 m instead of 3.8 x 10^8  ?

Radius of Moon's orbit IS the distance between the centres of the two planets
You don't need to add the Earth's radius to that

[knightrider]

Radius of Moon's orbit IS the distance between the centres of the two planets
You don't need to add the Earth's radius to that

Thanks for clarifying  !!

[Maz]

Just wanted to say thanks to lzxnl and Syndicate for answering my question...it's not letting me repost the quote
but thanks guys 

[Maz]

Hey...a lot of questions these days- physics is so not my thing though i wish it was
Can someone please tell me how to do this one? please- would greatly appreciate it 
A girl is swinging on a maypole
The girl has mass of 36kg and when she is moving with a speed of 2ms^1 the light rope makes an angle of 20 degrees with the vertical. Consider the motion of the centre of mass of the girl, which moves in a horizontal circle of radius r.
1. what are the vertical and horizontal components of tension? i think vertical is Tcos30=mv^{^2}/r....(36*2)/r= Tcos30
which will mean 76.62=Tr...and just T would be (76.62/r)...but i don't know if thats right...or how to do the horizontal tension
2. What is the tension in the rope
3. What is the net force acting on the girl and the radius of the circle

Thankyou so much in advance

[Shinkaze]

Can anyone help me with one of these super basic physics questions that I still don't know how to do .-. thanks!

A car travelling with a constant speed of 80 km/h passes a stationary motorcycle policeman. The policeman sets off in pursuit, accelerating uniformly to 80 km/h in 10.0 s and reaching a constant speed of 100 km/h after a further 5.0 s. At what time will the policeman catch up with the car?

[TheAspiringDoc]

Does the sun convert hydrogen nuclei into helium nuclei?
And if so, why does this process release 'pure energy'?
Thanks!

[happymeal]

Can anyone help me with one of these super basic physics questions that I still don't know how to do .-. thanks!

A car travelling with a constant speed of 80 km/h passes a stationary motorcycle policeman. The policeman sets off in pursuit, accelerating uniformly to 80 km/h in 10.0 s and reaching a constant speed of 100 km/h after a further 5.0 s. At what time will the policeman catch up with the car?

To do this, the first step you need to do is construct a velocity-time graph.
You want to depict both the policeman and the car on it (remember to convert km/h to m/s, you can do this by dividing km/h by 3.6)
Your graph should show the policeman accelerating to the same speed as the car in 10seconds, and then rising above it in the next five seconds.
Next, you want to tackle the distance that the police car has moved first. You do this by calculating the area of the triangle under the graph (10 seconds) and the trapezium under the graph (from 10 to 15 seconds). The police car now continues at a constant speed, so the formula you should have is x=(distance triangle)+(distance trapezium)+(speed times t-15). The t-15 part takes into account the distance travelled over the past 15 seconds.
Next, is the distance that the car travels. Throughout all this time, it travels at a constant velocity, so you can use the formula x=vt, subbing in your value for v.

Now to find when they meet, you equate the two equations together, giving you (distance triangle)+(distance trapezium)+(speed times t-15)=vt. The only unknown variable you should have here is t, which you will find using algebra.
Let me know if you have any questions, I tried to go through each step and it got pretty long in the end.

[lzxnl]

Does the sun convert hydrogen nuclei into helium nuclei?
And if so, why does this process release 'pure energy'?
Thanks!

Yes. It's called (not surprisingly) the 'proton-proton process'
I won't go into the details, but this process releases energy because of the strong nuclear force between protons and neutrons (if you ever wondered how positive charges could be so close together in the nucleus, it's because of this strong force that is REALLY strong at small distances but drops off really fast at higer distances)

This attraction releases energy. It's a bit like putting two opposite-pole magnets near each other; their opposite attraction will lead to an increase in kinetic energy.

[TheAspiringDoc]

Yes. It's called (not surprisingly) the 'proton-proton process'
I won't go into the details, but this process releases energy because of the strong nuclear force between protons and neutrons (if you ever wondered how positive charges could be so close together in the nucleus, it's because of this strong force that is REALLY strong at small distances but drops off really fast at higer distances)

This attraction releases energy. It's a bit like putting two opposite-pole magnets near each other; their opposite attraction will lead to an increase in kinetic energy.

thank you!


I'm a little unsure about Red Giant formation -  the contraction of a main sequence star (due to gravity) heats up the MSS (I get that). This causes Helium atoms to fuse together, whilst H fusion continues elsewhere. (I get all that too). But then, how does the He fusion (and H fusion) cause the main sequence star to expand (resultin in a red giant)??

[mohakmalhotra]

Hi there,
              i am not sure whether i have got the right answer for this question. It would be very nice if I could get your answers for this question. Thanks in advance

 Question : Theon is firing arrows from the ground towards targets mounted on a wall 6 m above the ground. He fires his arrows at a speed of 70 m/s and at an angle of 60°.
a.   What is the maximum height above the ground the arrows reach? (2 marks)
b.   How long does it take the arrow to reach the target? (3 marks)
c.   What horizontal distance is the target away from Theon? (2 marks)

[Syndicate]

Hi there,
              i am not sure whether i have got the right answer for this question. It would be very nice if I could get your answers for this question. Thanks in advance

 Question : Theon is firing arrows from the ground towards targets mounted on a wall 6 m above the ground. He fires his arrows at a speed of 70 m/s and at an angle of 60°.
a.   What is the maximum height above the ground the arrows reach? (2 marks)
b.   How long does it take the arrow to reach the target? (3 marks)
c.   What horizontal distance is the target away from Theon? (2 marks)

a. 193.5 m
b. 112.37 s
c. 433.01 m

If you want an explanation with how I derived with these answers, I can surely help 

[TheAspiringDoc]

Hi
few more astronomy questions!

What is the nature of cosmic microwave background radiation?
A)faint uniform x-ray glow
B)bright uniform x-ray glow
C)faint uniform radio signal
D)weak and patchy glow at visible wavelengths

The major cause of a supernova is:
A)gravitational collapse of the core of a massive star
B)exhaustion of nuclear fuel

Any help would be really appreciated as I have a test tomorrow!