<span>I am pretty sure the correct answer is C.
It makes the most sense.
Please correct me if I'm wrong.</span>
Answer: a) Moment = 200N.m
b) F = 500N
Explanation: <u>Moment</u> or <u>Torque</u> is the tendency of a body to rotate because of an aplplied force. It can be calculated as

F is force applied
r is distance from the force to the point of rotation
θ is the angle formed between force and distance
a) Since F and r are perpendicular, θ=90°, sin90° = 1, so


200
<u>Torque</u><u> caused by weight of the barrier at the pivot is </u><u>200N.m</u>
<u />
b) Counterweight has the same torque as pivot:
200
So, to find Force necessary to balance the weight:



F = 500
<u>It is necessary </u><u>a force of 500N</u><u> from the counterweight to balance the weight.</u>
<u />
A. 75 ampere-turns
<u>Explanation:</u>
Given:
Number of turns, n = 50
Voltage, V = 9V
Current, I = 1.5A
MMF= ?
MMF is magnetomotive force.
MMF is independent on voltage and dependent on the number of turns and current in the wire.
Thus, MMF can be written as:

On substituting the value we get:
MMF = 50 X 1.5
MMF = 75 ampere-turns
Answer:
The value is 
Explanation:
From the question we are told that
The volume of the bottle is 
The gauge pressure of the air is 
Generally the volume of air before the bottle is turned upside down is



Generally the volume air when the bottle is turned upside-down is


From the the mathematical relation of adiabatic process we have that

Here r is a constant with a value r = 1.4
So


The frequency of the human ear canal is 2.92 kHz.
Explanation:
As the ear canal is like a tube with open at one end, the wavelength of sound passing through this tube will propagate 4 times its length of the tube. So wavelength of the sound wave will be equal to four times the length of the tube. Then the frequency can be easily determined by finding the ratio of velocity of sound to wavelength. As the velocity of sound is given as 339 m/s, then the wavelength of the sound wave propagating through the ear canal is
Wavelength=4*Length of the ear canal
As length of the ear canal is given as 2.9 cm, it should be converted into meter as follows:

Then the frequency is determined as
f=c/λ=339/0.116=2922 Hz=2.92 kHz.
So, the frequency of the human ear canal is 2.92 kHz.