Two 6.0-cm-diameter electrodes are spaced 5.0mm apart. they are charged by transferring 1.1×10,,electrons from one electrode to the other. an electron is released from rest at the surface of the negative electrode. (a) how long does it take the electron to cross to the positive electrode? (b) what is its speed as it collides with the positive electrode? assume the space between the electrodes is a vacuum.

a) 20.11 μs; b) 248.6 m/s

Explanation: In order to solve this problem we have to use the New law to calculate the acceleration given by the electric force between the plates.

Also we know that E=Q/(A*εo) where A is the cross section of the plates

E=11*1,6*10^-19/(π*0.03^2*8.85*10^-12)=70.33 N/C

Then we have;

F=e*E= m*a so

a=e*E/m=12.36 *10^6 m/s^2

Finally to calculate the time, we use the kinematic equations ;

xf=(a*t^2)/2=

t= (2*xf/a)^1/2= (2*2.5*10^-3/12.36*10^6)^1/2 =20.11 μ s

then the velocity is calculate from:

vf= a*t= 12.36 *10^6* 20.11*10^-6=248.6 m/s

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form