Answer:
The ballon will brust at
<em>Pmax = 518 Torr ≈ 0.687 Atm </em>
<em />
<em />
Explanation:
Hello!
To solve this problem we are going to use the ideal gass law
PV = nRT
Where n (number of moles) and R are constants (in the present case)
Therefore, we can relate to thermodynamic states with their respective pressure, volume and temperature.
--- (*)
Our initial state is:
P1 = 754 torr
V1 = 3.1 L
T1 = 294 K
If we consider the final state at which the ballon will explode, then:
P2 = Pmax
V2 = Vmax
T2 = 273 K
We also know that the maximum surface area is: 1257 cm^2
If we consider a spherical ballon, we can obtain the maximum radius:

Rmax = 10.001 cm
Therefore, the max volume will be:

Vmax = 4 190.05 cm^3 = 4.19 L
Now, from (*)

Therefore:
Pmax= P1 * (0.687)
That is:
Pmax = 518 Torr
63783626736377474737377447
So my opinion that is false bc. the psychological profiling not is necessary being as criminal profiling ,the criminal mean a greater,intensive word in the way of ,,killer" ,when the ,,psychological" profile is total different way
hope helped
Answer:
.
Explanation:
The frequency
of a wave is equal to the number of wave cycles that go through a point on its path in unit time (where "unit time" is typically equal to one second.)
The wave in this question travels at a speed of
. In other words, the wave would have traveled
in each second. Consider a point on the path of this wave. If a peak was initially at that point, in one second that peak would be
How many wave cycles can fit into that
? The wavelength of this wave
gives the length of one wave cycle. Therefore:
.
That is: there are
wave cycles in
of this wave.
On the other hand, Because that
of this wave goes through that point in each second, that
wave cycles will go through that point in the same amount of time. Hence, the frequency of this wave would be
Because one wave cycle per second is equivalent to one Hertz, the frequency of this wave can be written as:
.
The calculations above can be expressed with the formula:
,
where
represents the speed of this wave, and
represents the wavelength of this wave.