if you were to draw a 3rd harmonic of a tube open at both ends, what would you draw at the ends of the tube?
1 answer:
A 3rd harmonic of a tube open at both ends will have displacement antinodes at both ends.
In a tube of length L with two open ends, the longest standing wave has displacement antinodes (pressure nodes) at both ends. The fundamental or first harmonic is what it is known as. The second harmonic is the longest standing wave in a tube of length L with two open ends.
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Answer:
Vf = 3.67 [m/s]
Explanation:
To solve this problem we must use the following equation of kinematics.
where:
Vf = final velocity [m/s]
Vi = initial velocity = 4.3 [m/s]
a = acceleration or desacceleration = 0.5 [m/s²]
x = distance = 5 [m]
Note: The negative sign in the above equation means that the velocity of the ball is decreasing (desacceleration).
Now replacing:
Vf² = (4.3)² - (2*0.5*5)
Vf² = 18.49 - 5
Vf² = 13.49
using the square root, we have.
Vf = 3.67 [m/s]
Answer:
Tension of 132N
Explanation:
We need to apply Summatory of Force to find the tension in the hand.
We define te tensión in the hand as and the Tension in fence post as
, then
We apply summatory of moments then
Where the Force 2 is 1.25m from the center of summatory,
We can note that,
We have two equation and two incognites, then replacing (1) in (2)
Answer:
mu=12Tm^2
Explanation:
the magnetic moment mu of a single loop is given by:
where I is the current, B is the magnetic field and A is the area of the loop. By replacing we obtain:
hope this helps!!