Rishi's Physics Thread

[Rishi97]

If the frictional force cannot exceed 8000N, then the centripetal force = 8000N

F = mv^2 / r
8000 * 5 = 400 * v^2
v^2 = 100
v = 10m/s

Is the answer right? That's what I got anyway

Yes that's right Yacoubb. Thanks

[Yacoubb]

Yes that's right Yacoubb. Thanks

No worries .

[Rishi97]

One of Jupiter's moons is Callisto
Callisto:
Distance from the centre of Jupiter (m) = 1.87 x 10⁹
Mass (relative units) = 2.6

What is the period of rotation of callisto? In days? (Answer is 16 days)

There is also info about other moons but I didn't think it was important so I haven't written it all here.

Thanks

[Stevensmay]

Mass (relative units) = 2.6

The question is saying that Callisto is 2.6 times heavier than Jupiter?
Unless I've missed an equation (likely) then we also need an additional piece of data, maybe the mass of Jupiter or it's period of rotation.

[Rishi97]

The question is saying that Callisto is 2.6 times heavier than Jupiter?
Unless I've missed an equation (likely) then we also need an additional piece of data, maybe the mass of Jupiter or it's period of rotation.

I'm not sure, it just says mass in relative units. I got confused as well cause the formula requries kg

[Thorium]

One of Jupiter's moons is Callisto
Callisto:
Distance from the centre of Jupiter (m) = 1.87 x 10⁹
Mass (relative units) = 2.6

What is the period of rotation of callisto? In days? (Answer is 16 days)

There is also info about other moons but I didn't think it was important so I haven't written it all here.

Thanks

When I looked at this question, I was sure I had done this before. So I found it on the checkpoints. It was also required for u too mention the distance and period of atleast one other moon. So take Europa, it has distance from centre of Jupiter 6.7x10^8 m and period of 3.55 days.

Now we can use the given info to find the period of rotation of callisto using one of kepler's law: "R^3/T^2=GM/4pi^2 for all satellites of a central mass."
So we can say that:
(6.7x10^8)^3/(3.55x24x3600)^2=(1.87x10^9)^3/T^2
..... T=1.4x10^6s=16.6 days

Not sure why it is 17 days if we round up. But I checked it with other moons and still got around 16.6.

[Rishi97]

When I looked at this question, I was sure I had done this before. So I found it on the checkpoints. It was also required for u too mention the distance and period of atleast one other moon. So take Europa, it has distance from centre of Jupiter 6.7x10^8 m and period of 3.55 days.

Now we can use the given info to find the period of rotation of callisto using one of kepler's law: "R^3/T^2=GM/4pi^2 for all satellites of a central mass."
So we can say that:
(6.7x10^8)^3/(3.55x24x3600)^2=(1.87x10^9)^3/T^2
..... T=1.4x10^6s=16.6 days

Not sure why it is 17 days if we round up. But I checked it with other moons and still got around 16.6.

Thanks Thorium... haha yeah, it's from checkpoints. Ok so the other info was necessary?
I don't think we went into Kepler's equation in much detail.

Thanks again

[Rishi97]

Can someone please explain the difference between photoconductive and photovoltaic mode?
Thanks in advance

[rhinwarr]

Photoconductive mode is when the photodiode is in reverse bias. If you look at a typical graph, you can see that it allows a small amount of 'dark current' through, depending on the light intensity. This mode is used primarily for varying the current in the circuit depending on light intensity. A photodiode in reverse bias is more sensitive than an LDR but less sensitive than a phototransistor. However it has the quickest response time.
Photovoltaic mode is when the photodiode is in forward bias. This is for converting light energy into electrical energy, thus powering a circuit. An example would be solar panels.

[Rishi97]

If resistance increases, does voltage increase or decrease? And why?

[rhinwarr]

That's not a very specific question but from the formula V=IR, if resistance increases you would expect voltage to also increase.

[lzxnl]

If resistance increases, does voltage increase or decrease? And why?

Generally if you're talking about a resistor in series, then yes, the voltage will increase. It's not just because of the formula V=IR.

Consider a circuit with a certain resistor of resistance R and let the rest of the circuit have resistance R', for a total effective resistance R+R'. Then, the current is V/(R+R'), so the voltage over the resistor R is RV/(R+R') = V*(1-R')/(R+R'). Here, as R' is a constant, if we increase R', we decrease the value of the fraction to be subtracted, thus increasing the voltage of the resistor. THAT is why voltage normally increases.

However, if you have resistors in parallel, the resistance will not affect the voltage over the resistor (if it's the only resistor and no other resistor is in series with it).

[Rishi97]

Explain why a magnet will always try to align itself with the direction of the existing magnetic field

Thanks

[lzxnl]

This goes beyond the scope of the course, but it can be shown that the potential energy of a magnet in a magnetic field is given by the negative dot product of the dipole moment u and the magnetic field B. The dot product is largest when the vectors are parallel, so as it's a negative dot product, the potential energy is lowest when the dipole moment (which is directed from the south to the north pole) is parallel to the magnetic field.

[Rishi97]

Christine is bouncing on her trampoline. At her maximum bounce height the bottom of her feet are 1.5 m above the level of the unstretched trampoline surface. At the lowest point of her bounce the bottom of her feet are 0.4m below the unstretched level of the trampoline surface. Christine weighs a bouncy 48 kg
From this information, what is the best estimate of the spring constant of the trampoline?