Recent content by A13235378

And how would the reaction be with the diluted acid?

One of the alternatives was NaBr My resolution was as follows: NaBr + HCl forms NaCl and HBr I researched and saw that HBr would be a gas and would be denser than air because its molar mass is greater. However, the answer was not that. My error is in which step? In the reaction? What would...

Yes you are right. I already found my mistake

Exactly, they are mesocompounds, but they also count as stereoisomers.

Summary:: Isomery My attemption: (I found 5) But the answer is 3 forms, where I missing?

Before, sorry for my English, it is not my native language I already have the solution to the issue, I just didn't understand a step. 1) Calculation of weight force (vertical): $$ F_v = \frac{\rho \pi R^2 . L .g}{4}$$ 2) Calculation of force due to water pressure on the plate (horizontal)...

We know that both the interior and the surface of an electrostatically balanced conductor are equipotential. My question is if when we approach the loaded objects, the surface of the conductor will continue to be an equipotential. If not, then there could be a field line that left the region...

Sorry for my concordance errors. English is not my native language, so I need to use a translator. I hope the drawing improves what I mean. Since the plate is positively charged, and we have a negative charge, the particle will be attracted. The field inside has already been given, varying with...

My doubt is in the mathematics part. As the force will be contrary to the direction of the field due to the negative charge, the particle will be attracted to the origin, so I assumed that the speed will increase (this is also found in the exercise response). I then applied an integral for the...

I traced a spherical X-ray Gaussian (green) where the negative charges were diametrically opposite. My question is this: I can transform the entire charge of the Gaussian sphere into a point charge placed in the center. So, can I analyze only the electrical forces of the two negative charges...

My attempt, Considering that it jumps in the maximum compression position: $$\frac{kA^2}{2} = mg(H+A)$$ replacing k / m with w ^ 2 : $$A^2 w^2-2gA-2gH=0$$ Solving the second degree equation: $$A=\frac{2g+\sqrt{4g^2+8gHw^2}}{2w^2}$$ But the answer is...

Nothing

but at what point will the normal force be zero?

Attempted solution: Consider the instant when the normal force of the lower ball is zero. Conserving energy: $$\frac{mv^2}{2}+mgh_1=\frac{mv_1^2}{2}+\frac{mv_2^2}{2} + mgh_2$$ Applying the resulting torque to the upper ball where the rotation point is the lower ball. $$T=I.a =...

First, I calculated the acceleration of A by the slope: -0.5m / s ^ 2 in t=4/s V=Vo+at (To bodie A) V=4-4.0,5 V=2m/s With that, I was able to find the acceleration of B: 0,5m/s^2SA= 10 +4t - t^2/4 SB= t^2/4 I equaled the two, finding t = 10s so, Va=-1m/s and Vb = 5m/s --> Vb,a = 6...