Answer:
Explanation:
Two frequencies with magnitude 150 Hz and 750 Hz are given
For Pipe open at both sides
fundamental frequency is 150 Hz as it is smaller
frequency of pipe is given by
where L=length of Pipe
v=velocity of sound
for n=1
and f=750 is for n=5
thus there are three resonance frequencies for n=2,3 and 4
For Pipe closed at one end
frequency is given by
for n=0
for n=2
Thus there is one additional resonance corresponding to n=1 , between and
Answer:
B.
Explanation:
One of the ways to address this issue is through the options given by the statement. The concepts related to the continuity equation and the Bernoulli equation.
Through these two equations it is possible to observe the behavior of the fluid, specifically the velocity at a constant height.
By definition the equation of continuity is,
In the problem is
, then
<em>Here we can conclude that by means of the continuity when increasing the Area, a decrease will be obtained - in the diminished times in the area - in the speed.</em>
For the particular case of Bernoulli we have to
For the previous definition we can now replace,
<em>Expressed from Bernoulli's equation we can identify that the greater the change that exists in pressure, fluid velocity will tend to decrease</em>
The correct answer is B: "If we increase A2 then by the continuity equation the speed of the fluid should decrease. Bernoulli's equation then shows that if the velocity of the fluid decreases (at constant height conditions) then the pressure of the fluid should increase"
Answer:
B_o = 1.013μT
Explanation:
To find B_o you take into account the formula for the emf:
where you used that A (area of the loop) is constant, an also the angle between the direction of B and the normal to A.
By applying the derivative you obtain:
when the emf is maximum the angle between B and the normal to A is zero, that is, cosθ = 1 or -1. Furthermore the cos function is 1 or -1. Hence:
hence, B_o = 1.013μT