Recent content by hraghav

So my arrow marking location would be an appropriate representation of the direction since the star does travel upwards as it gets dimmer and towards the right as it gets redder?

What would the coordinates be for the initial and final positions of B-V and mv? I know that initial position of B-V is 0.333 but how would I find its final position since I do not know what (B-V)not is. Also as -1 is not on the scale of mv how would I choose the coordinates for mv?

Homework Statement: Effects of extinction in the Colour- Magnitude Diagram (CMD) of NGC 5139; Since extinction affects the position of a star on both the x-axis and y-axis, then draw an arrow that shows the direction a star moves on the CMD for Av = 1.0 Relevant Equations: No equations are...

Number of neutrinos detected per kg of water = 12/3*10^6 kg = 4*10^-6 neutrinos Neutrinos detected by 1 eyeball = 4*10^-6 neutrinos * 0.01kg = 4*10^-8 neutrinos I am not sure how to scale this to get 400 neutrinos. I did try multiple ways but either ended up with 3.69*10^9 or 4*10^-8 neutrinos...

TL;DR Summary: For a transiting exoplanet, we find it takes 4.3 days from the start of the transit for the host star to reach a minimum brightness, which lasts for 10 days. Show that the radius of the exoplanet is about 1/2 of Earth’s radius if its orbital radial velocity is 17 m/s. I am...

I found the correct answer for solid odium ie 8.157 kg but I am still struggling with the liquid mass. I also tried 14.69 - 8.157 = 6.533 for liquid but this is still not correct. Could someone please spot the error for me. Thanks

Oh is it possible if you could help me with the algebraic method of finding the answers as my finals is in the pen paper form and I would not be able to use this. This is one of the sample final practice questions

Qmelt, honorite = (m honorite)⋅(Lfusion, honorite) Qmelt, honorite = 4.51kg⋅15500J/kg = 69905J Qcool, odium = (m odium)⋅(Cliquid, odium)⋅(391.14K−331.44K) Qcool, odium = 14.69kg⋅431.1J/kgK⋅(391.14K−331.44K) Qcool, odium = 378071.6823 Qfreeze, odium = (m odium)⋅(L fusion, odium) Qfreeze, odium...

I am not sure about this but this is what I got: Qhonorite(Teq) = (mhonorite)⋅(Csolid_honorite)⋅(Tmelt_honorite − Tinitial_honorite) + (mhonorite) ⋅ (Lfusion_honorite) + (mhonorite)⋅(Cliquid_honorite)⋅(Teq−Tmelt_honorite)

I did try but it doesn't seem to work. Could you help me out with the steps for this as I am trying this method for the first time

Mass of honorite initially: 4.51 kg at 261.2 K Additional mass of honorite added: 3.75 kg at 261.2 K total mass of honorite is 4.51 kg + 3.75 kg = 8.26 kg Mass of liquid odium: 14.69 kg at 391.14 K Q melt = m honorite × L honorite Qmelt = 8.26kg × 15500J/kg Qmelt = 128030 J Q cool = (m odium)...

Yes that makes sense. Thanks I got it now

Q1 = 1.49kg⋅256.4J/kgK⋅(1108.1−327.7)K Q1 = 298140.8944 Q2 = 1.49kg⋅15139J/kg = 22557.11J Q3 = 1.49kg⋅221.6J/kgK⋅(327.7−56.4)K Q3 = 89578.9192 Q4 = 2.44kg⋅524.8J/kgK⋅(308−33.6)K Q4 = 351372.4928 Q5 = 2.44kg⋅58350J/kg = 142374J QL,total = Q1 + Q2 + Q3 QL,total = 298140.8944 + 22557.11 +...

Cooling Lerasium to its melting point: Q1 =m L⋅c L,liquid ⋅(T L,i −T L,melt) Q1 = 1.49kg⋅221.6J/kgK⋅(1108.1−56.4)K Q1 = 330.184 ⋅(1051.7) = 347254.512 J Heating Atium to its melting point: Q2 =mA⋅cA,solid ⋅(TA,melt −TA,i) Q2 = 2.44kg⋅524.8J/kgK⋅(308−33.6)K Q2 = 1280.512⋅(274.4) = 351372.493 J...

Yes I understood what you mean, Thanks I got the correct answer