Answer: a) when the cross section is doubled the resultant resistence is a half. This means the thicker wire have half resistence than the thinner wire.
Explanation: In order to explain this behaviur we have to consider the expresion for the resistence which is given by:
where L and A are the length and the cross section for the wire, respectively.
From this expresssion we can conclude the above, this means
R=εo*L/A if A is now 2A we have
R' = εo*L/2*A= R/2
-- Radio waves. (To hear the weather while I'm waking up.)
-- Light waves. (To see where the dog is so I don't step on him.)
-- Infrared waves. (To make my toast for breakfast.)
-- Microwaves. (To heat my oatmeal for breakfast. Also when I go in to my job as a microwave communications engineer.)
Answer:
I believe it's called rolling friction
<h2>
Answer: The spreading of waves behind an aperture ismore for long wavelengths and less for short wavelengths</h2>
Here we are talking about Diffraction and, in fact, waves diffract the most when their wavelength is about the same size of the gap or aperture.
Diffraction happens when a wave (mechanical or electromagnetic wave) meets an obstacle or a slit .When this occurs, <u>the wave bends around the corners of the obstacle or passes through the opening of the slit that acts as an obstacle, forming multiple patterns with the shape of the aperture of the slit.
</u>
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Note that the principal condition for the occurrence of this phenomena is that the obstacle must be comparable in size (similar size) to the size of the wavelength.
In other words, when the gap (or slit) size is larger than the wavelength, the wave passes through the gap and does not spread out much on the other side, but when the gap size is equal to the wavelength, maximum diffraction occurs and the waves spread out greatly.
This means the smaller the slit or obstacle (diffracting object), the wider the resulting diffraction pattern, and the greater the obstacle, the narrower de resulting patter.
Answer:

Explanation:
Let
be % of
at time
(time in minutes).
Amount of
in room=Volume of room 
Rate of change of
in room=Volume of room x 

Rate of inflow of
=

Rate of
outflow

Therefore rate of change of
is 
% rate of change is
Therefore we have:

Integrate both:


In
In

Raise both sides to base
,

Initially, there's only 0.25% of
,
Now substitute
and 

