Recent content by himura137

Three resistors are connected in series across a battery. The value of each resistance and its maximum power rating are as follows: 4.7 ohms and 17.7 W, 23.1ohms and 13.9 W, and 12.8ohms and 12.9 W. (a) What is the greatest voltage that the battery can have without one of the resistors burning...

1)Which of the following is not both a Bronsted-Lowry acid and a Bronsted-Lowry base? a. OH- b. HSO4- c. SH- d. HCO3- e. H2PO4- I knwo its not HCO3, H2PO4- AND HSO4-, but i cannot figure out whether OH can act as an acid and base or not, i know its produced when a base react...

thanks i got it, how did know that q(delta)V = 1/2mv^2, i was tryin to look for a relationship between velocity and potential but could not figure it out and can you check my work for #2, i stil could not get the correct answer

1) i am confused, i though the particle was at rest at A and V and arrive at B with some velocity and how did you know that potential at A and C (equal) is greater than B 2) i change the equation and got thsi i placed the charges as follow q1 q2 q 3 q4 and used q1 as the starting point...

1) The potential at location A is 441 V. A positively charged particle is released there from rest and arrives at location B with a speed vB. The potential at location C is 805 V, and when released from rest from this spot, the particle arrives at B with twice the speed it previously had, or 2...

Fx= 0 = (k|q1x||q2x|/ 1^2 ) + (k|q1x||q3x|/ 1^2 ) - (k|q1x||q4x|/ (*root*2)^2 )(cos 45) Fy= 0 = (k|q1y||q2y|/ 1^2 ) + (k|q1y||q3y|/ 1^2 ) - (k|q1y||q4y|/ (*root*2)^2 )(sin 45) so i actually found the force of q4x and q4y, which is same as q1x and q1y can you check #2 and #3 for me...

*bumped* i showed my work, can someone please check

#2 i used the V= Vo +at to find the acceleration of the proton, then multiply that by the mass of proton (1.67262 x 10^-27 kg) to get the Force, than used the equation E = F/q where q is the charge on an electron and since its a proton its positive with the unit (1.6022×10-19 C) and i get...

ok i got the answer i placed the charged like this q2 q4 q1 q3 i split the question into x and y and got Fx = (k|q1x||q3x|/ 1^2 ) - (k|q1x||q4x|/ (*root*2)^2 )(cos 45) Fy = (k|q1y||q2y|/ 1^2 ) - (k|q1y||q4y|/ (*root*2)^2 )(sin 45) and i get Fx and Fy as 2.54558 and that is the...

i put one of the q as q1 and the -0.9uc as q2 and q3 and the other q as q4 and used the equation F= (k|q1||q2| / r2) +(k|q1||q3| / r2) + (k|q1||q4| / r2) and since the F = 0 on the q and all of them contain k and q1, i just divide it straight through and got ( |q2| + |q3| ) 2 = q which when i...

why plant become dominately sporophyte rather than gametophyte? i am currently learning about mosses, fern, gymnosperm and angiosperms and i got stuck when trying to make a connection between them.

Hi, I am currently stuck on the following questions. Any help is greatly appreciated. thank you *all question are algebra based 1) A point charge of -0.90 microC is fixed to one corner of a square. An identical charge is fixed to the diagonally opposite corner. A point charge q is fixed to...