Recent content by ianc1339

"but it is possible to resolve into any two directions that are not collinear" "if you find the components of a vector in the directions of two vectors not at right angles then you cannot add the components together to recover the original vector" Aren't these two statements contradicting?

The answer is D (60 degrees) and I understand how to get that answer. But this assumes that the new velocity's component of v/4 can form right angles with another component of the new velocity. So I'm confused whether vector components always form right angles to each other. When I searched...

The first ionization energy decreases between group 5 and group 6 due to the repulsion between the electrons in the p orbital. Although I understand that the effective nuclear charge increases between group 1 and group 2 elements, why isn't this the case between group 1 and group 2 elements...

Oh, so the partial pressure of nitrogen is: ##0.75\times 7.4\times 10^4\times\frac{0.4}{0.75}=2.96\times 10^4 Nm^{-2}## And since volume is proportional to number of moles, this new mole fraction can be established using the volume percentages?

Does that mean the partial pressures of the gases decrease by 25% ? So do I do: ##N_2: 3.0\times 10^4\times 0.75=2.25\times 10^4 Nm^{-2}\\## ##O_2: 2.6\times 10^4\times 0.75=1.95\times 10^4 Nm^{-2}\\## I don't have the answers with me so I don't know if my answer is correct

Oh, so does the partial pressure of the other gases still remain constant even if carbon dioxide was removed?

It doesn't say anything about it so I'm assuming it to be constant volume.

I've first calculated the partial pressures of each gas: ##N_2: 0.4\times 7.4\times 10^4=3.0\times 10^4 Nm^{-2}\\## ##O_2: 0.35\times 7.4\times 10^4=2.6\times 10^4 Nm^{-2}\\## ##CO_2: 0.25\times 7.4\times 10^4=1.9\times 10^4 Nm^{-2}\\## From here, I do not know how to continue. Could someone...

So the relation becomes ##\frac{1}{T}=\frac{R}{p}\times\rho## where ##\frac{R}{p}## is a constant. ##\frac{R}{p} = \frac{1}{T\rho} = 2.7377\times 10^{-3}##. Since the weight of the air inside the balloon before take off is ##23789.25N##, the density of the air inside the balloon is ##\rho =...

So the weight of the material is ##W=mg=800\times 9.81 = 7848N##. So the weight of the air inside must be ##31637.25 - 7848 = 23789.25N##. Since volume is proportional to temperature, as temperature increases, density decreases. So temperature is inversely proportional to temperature?

Is it ##\rho V g##?

First, I tried using the Archimedes principle and calculated the weight of the surrounding air displaced when taking off. ##W = 2500\times 1.29\times 9.81 = 31637.25 N## But then, I got stuck and do not know how to proceed from here on. I don't want the full solution yet but can I get some...

My book said the specific latent heat of ice was 330 Jkg^-1. But with a quick google search, it turns out that it is actually 330000 Jkg^-1. And yes, I ended up getting 32 degrees Celsius. I don't know why the book wrote down the constant wrong which is quite annoying. Anyways, thanks for the...

Calculating latent heat required to melt ice: ## Q=mL_f\\ Q=0.150kg\times 330Jkg^-1\\ Q=49.5 J ## Calculating change in temperature of water: ## Q=mc\Delta T\\ −49.5J=0.35kg\times 4200Jkg^{-1}K^{-1}\times (T_f −80^o C)\\ T_f=79.966^oC ## Calculating temperature when the two bodies of...

It didn't say anything about it. So I'm assuming it to be 0 degrees Celsius.