Question #95e73

Question

↳Redirected from "How does a vector quantity differ from a scalar quantity?"

1521 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-09T18:56:53

#sqrt(2)/2#

#sqrt((1/2)^2+(1/2)^2)#

First find out what #(1/2)^2# is:

#(1/2)^2 = 1/4#

So, we now know that the expression is now #sqrt(2/4)# which is the same as #sqrt(1/2)#.

Then, we rationalize the denominator as we cannot have a root as
the denominator:

#sqrt(1) = 1#, so the expression is now #1 /sqrt(2)#

Multiply both the numerator and the denominator by #sqrt(2)#

Denominator #= sqrt(2) * sqrt(2) = 2#

Numerator #= 1 * sqrt(2) = sqrt(2 )#

Hence,

#= sqrt(2)/2#

Answer 2 · 2017-03-09T19:30:24

The expression is equivalent to #sqrt(2)/2#.

The square of #1/2# is #1/4# because #(1/2)(1/2) = 1/(2*2) = 1/4#.

Therefore,

#sqrt(1/4 + 1/4)#

#sqrt(2/4)#

#sqrt(1/2)#

We can separate the radicals.

#sqrt(1)/sqrt(2)#

#1/sqrt(2)#

I would recommend you rationalize the denominators.

#1/sqrt(2) * sqrt(2)/sqrt(2)#

#sqrt(2)/2#

Hopefully this helps!