Be able to list the three compounds that are formed as products of highly exothermic
1 answer:
You might be interested in
Answer:
<em>Part A</em><em>:</em>
a) If the wavelength of the light is decreased the fringe spacing Δy will decrease.
<em>Part B</em><em>:</em>
b) If the spacing between the slits is decreased the fringe spacing Δy will increase.
<em>Part C</em><em>:</em>
a) If the distance to the screen is decreased the fringe spacing will decrease.
<em>Part D</em><em>:</em>
The dot in the center of fringe E is farther from the left slit than from the right slit.
Explanation:
In the double-slit experiment there is a clear contrast between the dark and bright fringes, that indicate destructive and constructive interference respectively, in the central peak and then is less so at either side.
The position of bright fringes in the screen where the pattern is formed can be calculated with
- m is the order number.
is the wavelength of the monochromatic light.
- L is the distance between the screen and the two slits.
- d is the distance between the slits.
- Part A: a) In the above equation for the position of bright fringes we can see that if the wavelength of the light
- Part B: b) In the above equation for the position of bright fringes we can see that if the spacing between the slits d is decreased the fringes are going to be wider apart. That means the fringe spacing Δy will increase.
- Part C: a) In the above equation we can see that if the distance to the screen L is decreased the fringes are going to be closer. That means the fringe spacing Δy will decrease.
- Part D: We are told that the central maximum is the fringe C that corresponds with m=0. That means that fringe E corresponds with the order number m=2 if we consider it to be the second maximum at the rigth of the central one. To calculate how much farther from the left slit than from the right slit is a dot located at the center of the fringe E in the screen we use the condition for constructive interference. That says that the path length difference Δr between rays coming from the left and right slit must be
We simply replace the values in that equation :
The dot in the center of fringe E is farther from the left slit than from the right slit.
Missing details: figure of the problem is attached.
We can solve the exercise by using Poiseuille's law. It says that, for a fluid in laminar flow inside a closed pipe,
where:
is the pressure difference between the two ends
is viscosity of the fluid
L is the length of the pipe
is the volumetric flow rate, with
being the section of the tube and
the velocity of the fluid
r is the radius of the pipe.
We can apply this law to the needle, and then calculating the pressure difference between point P and the end of the needle. For our problem, we have:
is the dynamic water viscosity at
L=4.0 cm=0.04 m
and r=1 mm=0.001 m
Using these data in the formula, we get:
However, this is the pressure difference between point P and the end of the needle. But the end of the needle is at atmosphere pressure, and therefore the gauge pressure (which has zero-reference against atmosphere pressure) at point P is exactly 3200 Pa.
