Answer:A)For open pipe
1stπ=484cm
2nd wavelength=161.3cm
3rd is=96.8cm
B)For closed pipe
1st wavelength=242cm
2nd=121cm
3rd wavelength=80.7cm
Explanation:
An open pipe produces a standing waves whose wavelength is related to it's length(length of pipe) by nL/4=π ,where n represent odd integers 1,2,3...
So the first note called fundamental note or 1st Harmonic is L=π/ 4,121=π/ 4
π= 121×4=484cm
Third harmonic,L=3π /4
So π= 4l/3=4×121/3=161.3cm
The 3rd wavelength is
Fifth harmonic=
π =4l/5=4×121/5=96.8cm
For a pipe open at one end and closed at the other also called closed pipe
Fundamental note is
2L/n= π where n=1,2,3....ie integers
First wavelength
π=2L=2×121=242cm
Second wavelength
π=L=121cm
Third is
π=2L/3=2×121/3=80.7cm
Answer:
Explanation:
This question appears incomplete because of the absence of the data been talked about in the question. However, there is a general ruling here and it can be applied to the data at hand.
If an increase in the distance of charges (let's denote with "d") causes the electric field strength (let's denote with"E") to increase, then the mathematical representation can be illustrated as d ∝ E (meaning distance of charge is directly proportional to electric field strength).
But if an increase in the distance of the charges causes the electric field strength to decrease, then the mathematical representation can be illustrated as d ∝ 1/E (meaning distance of charge is inversely proportional to electric field strength).
A scatterplot can also be used to determine this. If there is a positive correlation (correlation value is greater than zero but less than or equal to 1) on the graph, then it is illustrated as "d ∝ E" BUT if there is a negative correlation (correlation value is less than zero but greater than or equal to -1), then it can be illustrated as "d ∝ 1/E".
Answer:
0.20 G
Explanation:
The given parameters are;
The magnetic field strength of planet X = 0.70 G
The location of a probe in relation to the vertical wire = 50 m
The measurement of a probe placed 50 mm east of a vertical wire = 0.50 G
The location of the probe = 20 mm east of the vertical wire
Based on the given parameter, the wire is a current carrying conductor with a magnetic field acting tangent to the magnetic field lines.
Given that the measured Earth magnetic field reduces, we have that the magnetic field of the wire cancels the magnetic field of the Earth
As the probe moves closer to the wire the measured magnetic field in the southern direction reduces
When the probe is placed 20 mm from the wire, the magnetic field reduces to 20/50 × 0.50 G = 0.20G
Answer: Image for option 4
Explanation:
The fourth option is correct because there are three forces acting on the block:
which is the <u>Friction force</u> and acts against the movement of the block
the <u>Normal Force</u>, which is perpendicular (normal) to the surface of the rough ramp
the Weight of the block, which is the force due gravity acceleration and has an <em>x-component</em> and an <em>y-component.</em>
Sadly, I haven't read the chapter, and I can't see Table 3.
But I've seen this fun question before, so instead of reporting
the question for incomplete content, I'm going to answer it
and take your points.
a). one decaration
b). two kilomockingbirds
c). one microphone
d). one nanogoat
e). one examiner
By the way ... perhaps it hadn't occurred to you, but YOU could have
easily answered these completely on your own. All you'd need to do
is take a look at Table 3 in the latest chapter of your Physics book !