1. Li3N lithium nitride
AND
3.SO2 sulfur dioxide
The process you're fishing for is "polarization", but that's a
misleading description.
Polarization doesn't do anything to change the light waves.
It simply filters out (absorbs, as with a polarizing filter) the
light waves that aren't vibrating in the desired plane, and
allows only those that are to pass.
The intensity of a light beam is always reduced after
polarizing it, because much (most) of the original light
has been removed.
A laser light source may be thought of as an exception,
since everything coming out of the laser is polarized.
On question 30, that is a displacement- time graph (DT). On this type of graph the gradient is equal to the velocity. B has the steepest gradient, then A and finally C
Now velocity is a vector quantity so it has a direction and speed ( speed doesn't have a fixed direction.)
on the DT graph im going to assume that movement B is a positive velocity with A and C being negative.
So by ranking these: A is the most negative, C is the least negative and B has to be the greatest as it is the only positive velocity.
Q31, The same type of graph is present, by looking at the gradients we can rank the largest and smallest velocities- speeds in the case of the question.
i'll skip my working out as its the same as before:
C, B, A and then D
the same idea as on Q30 applies to Q31 part b,
D,C,B then A
Answer:
Approximately
and approximately
.
Explanation:
Let
and
denote the capacitance of these two capacitors.
When these two capacitors are connected in parallel, the combined capacitance will be the sum of
and
. (Think about how connecting these two capacitors in parallel is like adding to the total area of the capacitor plates. That would allow a greater amount of charge to be stored.)
.
On the other hand, when these two capacitors are connected in series, the combined capacitance should satisfy:
.
(Consider how connecting these two capacitors in series is similar to increasing the distance between the capacitor plates. The strength of the electric field (
) between these plates will become smaller. That translates to a smaller capacitance if the amount of charge stored
stays the same.)
The question states that:
, and
.
Let the capacitance of these two capacitors be
and
. The two equations will become:
.
From the first equation:
.
Hence, the
in the second equation here can be replaced with
. That equation would then become:
.
Solve for
:
.
.
.
Solve this quadratic equation for
:
or
.
Substitute back into the equation
for
:
In other words, these two capacitors have only one possible set of capacitances (even though the previous quadratic equation gave two distinct real roots.) The capacitances of the two capacitors would be approximately
and approximately
(both values are rounded to two significant digits.)
Answer:
<em>The </em><em>heat </em><em>capacity</em><em> </em><em>of </em><em>a </em><em>body </em><em>is </em><em>defined</em><em> </em><em>as </em><em>the </em><em>heat </em><em>required</em><em> </em><em>to </em><em>raise </em><em>it's </em><em>temperature</em><em> </em><em>by </em><em>me </em><em>degree </em><em>or </em><em>one </em><em>kelvin.</em><em>w</em><em>h</em><em>i</em><em>l</em><em>e</em><em> </em><em>specific</em><em> </em><em>heat </em><em>capacity </em><em>of </em><em>a </em><em>substance</em><em> </em><em>is </em><em>defined</em><em> </em><em>as </em><em>the </em><em>heat </em><em>required</em><em> to</em><em> </em><em>the </em><em>temperature</em><em> </em><em>of </em><em>a </em><em>unit </em><em>mass </em><em>of </em><em>it </em><em>through </em><em>one </em><em>degree </em><em>or </em><em>one </em><em>kelvin</em><em>.</em>
<em>I </em><em>hope </em><em>it </em><em>helps</em>