Recent content by Luxdot

So what would happen if the angle is decreased? I.e. smaller than the correct one for that latitude?

Yes, I am talking about the angle! So this has to do with the rotational axis of the earth?

I am curious how the height of the gnomon in a horizontal sundial is dependent on your location. Why does it need to be different high depending on where you are? Does it have to do with how high the sun is in the sky depending on the latitude you are on?

Hi! I am doing some simple observations of different light sources with a simple DIY spectroscope. When I look at a computer screen I see what I believe to be an emission spectrum due to the dark spectrum with emission lines on it. Is this correct? And why does a computer screen emit an...

How can the inlet water temp be higher than the one of the feeder water that is in the reactor tank of a BWR?

Thanks! So due to the vacuum, the seawater will leak into the condensor but the steam will not be leaked out? Will there be any other consequences? I'm aware that there is low radiation in the turbine so it won't be dangerous in that way. But is there a risk of some kind of explosion of some...

Hi! I'm a bit curious, what would happen if there would be a leakage in the condenser in a nuclear plant (PWR or BWR). As I understood it the stean in the turbine is expanded to about 4 kPa and I guess the pressure outside in the freshwater that is pumped into the condenser is at atmospheric...

So, since the equivalent dose is 0,40 mSv = 0,0004 Sv and the mass 1,2 kg, the absorbed dose is 0,00048 Gy (or J/kg). To get the total absorbed energy, wouldn't I have to divide the absorbed dose with the mass again to get rid of the kg? I.e. 0,00048 [J/kg] / 1,2 [kg] = 0,00040 [J]? This would...

I see! Since the energy absorbed per kg is 40x1-4 J, the total absorbed energy is 4x10-4x1,2 = 4,8x10-4 Gy = 0,48 mGy? Would then the number of photons be 4,8x10-4 [J/kg]/8*10-15 = 6x1010 photons?

I understand that logic fine, I would be able to buy 50 cans... But there is something else that doesn't add up for me. 50 keV is 8*10-15 joules, right? If the energy of one photon is 50 keV (8*10-15 J), the dose would be the energy (Dollars) times by mass (Hours), i.e. 8*10-15*1,2 = 9,6 *10-15...

What's confusing me is that for example if I have an equivalent dose of 1 Sv and a mass of 1,5 kg. Then the absorbed dose should be 1,5 Gy? But then I have multiplied J/kg with kg which should remove kg? Am I wrong when I'm scaling the absorbed dose to the mass? Or should the absorbed dose be 1...

So let´s say, hypothetically that I have a radiation energy (energy of one photon) of 10 keV (which is equal to 1,6*10-15 J) and an equivalent dose of 1 Sv and a mass of 1 kg. The absorbed dose is then 1 J/kg or 1 Gy (assuming Q=1). This is also the total energy when multiplied by the mass (in...

Yes I think so! But I'm still a bit confused on how I find the absorbed dose?

One is Joule and one is eV... My bad! So the total energy should be 0,48 mJ = about 3*1015 eV and the energy of one photon was 50 keV, so it makes more sense

That's the equivalent dose... To clarify, the first part is to find the absorbed dose. I did this by multiplying the equivalent dose (0,4 mSv) by the mass (1,2 kg) and got 0,48 mJ. Since the absorbed dose should be in Gray I'm a bit confused now since I made the error of thinking that I got the...