Physics: Post your doubts here!

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Can anyone explain this?

I know it's quite late, but I hope the explanation helps nonetheless.
The spring is moving vertically up and down, with equilibrium point being in the centre. At "lowest point of its motion" the mass has MAXIMUM displacement downwards. And at maximum displacement, we know that potential energy is maximum and kinetic energy is minimum(=0) therefore velocity=0 at lowest point and answer is D :)
 
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Is there any whatsapp group for this group and if yes can someone kindly add me over there
 
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How to solve part b?
Hi, firstly this is a physics forum and your question is chemistry-related. Please be mindful henceforth :)
As to your question, it's quite a memory based answer. You need to know that MgCl2 dissolves in water to give Mg(OH)2+HCl; Since one product is basic and the other is acidic, your best guess is pH 7.
in case of MgO, you need to know that it hardly dissolves/is slightly soluble. It is more of a surface reaction that occurs in which Mg(OH)2 is produced. As a result, you predict a high pH. Dont go till 14, Mg(OH)2 is not as strong of an alkali.
 
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Hi, firstly this is a physics forum and your question is chemistry-related. Please be mindful henceforth :)
As to your question, it's quite a memory based answer. You need to know that MgCl2 dissolves in water to give Mg(OH)2+HCl; Since one product is basic and the other is acidic, your best guess is pH 7.
in case of MgO, you need to know that it hardly dissolves/is slightly soluble. It is more of a surface reaction that occurs in which Mg(OH)2 is produced. As a result, you predict a high pH. Dont go till 14, Mg(OH)2 is not as strong of an alkali.
Sorry. I didn't notice at that time and thanks.
 
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View attachment 65219 Can someone please help me solve this problem?
To do that we need to use formula F=MA
Horizontal component will be ( 2m/s^2 * 0.05kg = 1N to the left
Vertical component are caused by the effect of gravitational pull, if Mechanics "10m/s^2" Physics "9,81m/s^2". Simply use 0.05kg * 10m/s^2 = 0.5N downwards.

Solving using Pythagoras theorem, tan-1( 1 / 0.5 ) 63.4 degrees
 
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Can anyone help me with this?? "B"View attachment 65228
Consider the weight of a column of air of cross-sectional area A and height h stretching from sea level to the edge of the atmosphere.
The density of the air decreases linearly with height above sea level, so the average density of air in the column is ρ/2 where ρ is the density of air at sea level.
The difference in pressure between the bottom and the top of the column of air is P, the atmospheric pressure at sea level.
The weight of the air column is ρgAh/2 and therefore P = ρgh/2

Solve for h (y)
 
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