Sriman Dutta

Ionised HCl is more active than unionised HCl. Ionisation makes acids capable to react readily, because acids get dissociated when dissolved in water.

You may have a skill for writing poetry. You may have a skill for solving mathematical puzzles. You can improve your skills by practice. And when practice becomes a passion, you innovate over your skill to create new things. Imagination, if logical, is also an innovation. In my words, technology is something similar. Technology is nothing but the crop grown from the seed of logical imagination.

From Fact 2 we know that the AE and AD are same, so altitude AE bisects BC. This implies that the triangle is either equilateral or isoceles.

Hi, paragaster are you an Indian? Or Hindu researcher?

In an inelastic collision the sum total of kinetic energies of two bodies does not equal to the total kinetic energy of the system after collision. For such a collision, the equation KE1+KE2=KE3 does not hold true. But momentum is conserved so that we get m1u1+m2u2=(m1+m2)v, where u1 and u2 are initial velocities and v is the final velocity of the whole system.

I am not trying to build a relationship between dissolution rate and drag. What I want is to get a model which shows how the drag forces change when a solid moving down gets dissolved in the liquid.

Thanks mordred for clarification.

What's the concept of duplicity?

India has also been a strong supporter of NAM(Non-Alignment Movement). She wanted to maintain peace in the world and avert the Cold War.

OK fine. It has been mentioned earlier that dissolution is not depended upon drag. In the equation, I haven't defined rate of dissolution with respect to drag. Instead the drag force depends upon the radius of the solid, which depends upon rate of dissolution and time. [math] r=\frac{dV}{dt}=\frac{4\pi}{3}\frac{dR^3}{dt}[/math]

If the solid dissolves in the liquid without undergoing any chemical reaction, why shall be there thermal energy? Is the energy coming from the fact that drag acts as a resistance to the downward motion of the solid? Moreover will there be any Magnus effect?

Can you explain the derivation?

Your defeat speaks about your deficiencies. From my experience, I can say that every time life comes to such a situation which proves that what you have known about yourself and your life is wrong. You have to learn it again and correct the wrongs that you have done.

America tried to cause an immediate end to World War II by attacking Japan in August 1945. But this intervention might not have been very good for world peace as this attack used nuclear weapons. America in the 21st century has emerged as a friend of India and an enemy - perhaps a sworn enemy of IS. But we need to see what America does under Trump, isn't it?

Hi, I got some equations.... Please see whether they are right or not. [math]Force on the solid in the downward direction = Weight of the solid - Drag force - Buoyant force[/math] [math]ma=mg-6\pi \mu Rv - V\rho_l g[/math] where [math]m[/math] is mass of the solid and [math]\rho_l[/math] is the density of the liquid [math]m=V\rho_s[/math], where [math]\rho_s[/math] is the solid's density and [math]V[/math] is its volume. Substituting this in the above equation, [math]V\rho_s a = V\rho_s g - 6\pi\mu Rv- V\rho_l g [/math], [math] v [/math] is the solid's downward velocity [math]\frac{4}{3}\pi R^3\rho_s a = \frac{4}{3}\pi R^3\rho_s g - 6\pi\mu Rv - \frac{4}{3}\pi R^3\rho_l g [/math] Since both sides are differentiable by [math]dt[/math], and using the definition [math]k=\frac{4}{3}\pi[/math], we get- [math]k\rho_s\frac{d(R^3a)}{dt}= k\rho_sg \frac{dR^3}{dt}-6\pi\mu\frac{d(Rv)}{dt}-k\rho_lg\frac{dR^3}{dt}[/math]

What test? Whose test? Why test? How test? Where test?

[math]KE=\frac{1}{2}mv^2=\frac{1}{2}pv=\frac{p^2}{2m}[/math]

LOL

Here's a simple demonstration of the law of coe. Take a ball of mass m and at height h above the ground. At that height: PE=mgh, KE=0, Total E=KE+PE=mgh At midway, when ball is at a height of x above ground level, velocity of ball(v)=sqrt(2g(h-x)) PE=mgx, KE=0.5mv^2=0.5m(2g(h-x))=mgh-mgx, Total E=KE+PE=mgh At the ground, velocity(v)=sqrt(2gh) PE=0(since height=0), KE=0.5mv^2=0.5m(2gh)=mgh, Total E=KE+PE=mgh

We are already living in a Machine Era. The time when humans marry robots is not too far away I think.

But one thing seriously disabled, The intelligent alien society might be living in a very far away place in the universe. Since their tech is more advanced they might come to earth through a wormhole.

Hello, Suppose there's a spherical solid of initial radius [math]R[/math]. This solid when dropped in a certain liquid dissolves at the rate [math]r[/math]. By rate of dissolution I mean the loss of the volume of the solid in unit time. By using derivatives, we can represent it as- [math]r=\frac{dV}{dt}=\frac{4\pi}{3}\frac{dR^3}{dt} [/math] where [math]dV[/math] is change in volume, which is directly proportional to the cube of the change in radius of the spherical solid represented as [math]dR^3[/math]. Since the solid moves downward it experiences drag. By Stoke's Law, drag is - [math]F_D=6\pi \mu Rv [/math] where [math]\mu[/math] is the dynamic viscosity of the liquid, [math]v[/math] is the solid's downward velocity. But, here downward velocity [math]v[/math] of the solid is a function of the drag force, which is again a function of the radius. If we apply Newton's Laws, [math]Force on the solid in the downward direction = Weight of solid - Drag force [/math] [math]ma=mg-F_D[/math] where [math]m[/math] is the solid's mass. But mass also depends upon radius. All the variables seem inter-related. I want to reduce the whole situation into an equation but cannot proceed beyond this. Please suggest what and how to proceed.

com=conservation of mass????

Both the conservation laws hold true. Besides, there's also the conversation of mass. In case of collisions between two bodies having their masses as [math]m_1[/math] and [math]m_2[/math] respectively and velocities before collision as [math]u_1[/math] and [math]u_2[/math] and after collision, as [math]v_1[/math] and [math]v_2[/math] respectively, then the following equation holds true - [math]m_1u_1+m_2u_2=m_1v_1+m_2v_2[/math]

Quite true. But, (since the thread is completely hypothetical) what if human race is artificially eradicated by an extraterrestrial alien community??