The electric field strength at any point from a charged particle is given by E = kq/r^2 and we can use this to calculate the field strength of the two fields individually at the midpoint.
The field strength at midway (r = 0.171/2 = 0.0885 m) for particle 1 is E = (8.99x10^9)(-1* 10^-7)/(0.0885)^2 = -7.041 N/C and the field strength at midway for particle 2 is E = (8.99x10^9)(5.98* 10^-7)/(0.0935)^2 = <span>-7.041 N/C
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Note the sign of the field for particle 1 is negative so this is attractive for a test charge whereas for particle 2 it is positive therefore their equal magnitudes will add to give the magnitude of the net field, 2*<span>7.041 N/C </span>= 14.082 N/C
Answer:
C- Equilateral triangle
Step-by-step explanation:
I am not very sure dear
Answer:
I think option A is correct answer
Answer:
x=40
Step-by-step explanation:
Because every triangle's angles add up to 180 degrees, you know 3x+x+20=180. Since 3x and x are like terms, you can add them to get 4x. 20 and 180 are also like terms. However, to move your 20 to the side of 180, you need to subtract it. So you have 4x=160. To get x by itself, you divide 4x and 160 by 4. 4x/4 is x, and 160/4 is 40. So your answer is x=40.
f(x)=(x+a)/b
or bf(x)=x+a
let f(x)=y
by=x+a
flip x and y
bx=y+a
or y=bx-a
or f^{-1}(x)=bx-a
also g(x) is inverse of f(x)
bx-a=cx-d
so b=c,a=d
again let g(x)=y
y=cx-d
flip x and y
x=cy-d
cy=x+d
y=(x+d)/c
or g^{-1}(x)=(x+d)/c
also f(x) is inverse of g(x)
so (x+a)/b=(x+d)/c
so a=d,b=c
so in either case a=d,b=c
take b=c=1
a=d=2
f(x)=(x+2)/1=x+2
g(x)=1x-2=x-2
so f(x) and g(x) are two parallel lines f(x) with y- intercept=1 and slope 0
g(x) with y-intercept -2 and slope 0
if we take b=c=2,a=d=3
f(x)=(x+3)/2=x/2+3/2
g(x)=2x-3
here f(x) is of slope 1/2 and y-intercept 3/2
g(x) is of slope 2 and y intercept -3
part 3.
f(f(x))=g((x+a)/b)=c[(x+a)/b]-d=(c/b)(x+a)-d
