One of the harmonics of a column of air in a tube that is open at one end and closed at the other has a frequency of 448 Hz, and
1 answer:
Answer:
1.34 m
Explanation:
For an open-end tube, the frequency difference between two consecutive harmonics is equal to the fundamental frequency of the tube:
In this case, we have
so, the fundamental frequency is
For an open-end tube, the fundamental frequency is also given by:
where v is the speed of sound and L the length of the tube.
Since we know v = 343 m/s, we can solve the formula for L:
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Answer:
for 4.567 I can't tell if it x or x
so:
answer using x = 0.0000644697N
answer using x = 0.00644697
Explanation:
use equation F = GMm/
Answer:
Velocity = 3.25[m/s]
Explanation:
This problem can be solved if we use the Bernoulli equation: In the attached image we can see the conditions of the water inside the container.
In point 1, (surface of the water) we have the atmospheric pressure and at point 2 the water is coming out also at atmospheric pressure, therefore this members in the Bernoulli equation could be cancelled.
The velocity in the point 1 is zero because we have this conditional statement "The water surface drops very slowly and its speed is approximately zero"
h2 is located at point 2 and it will be zero.
Answer:
6.53 m/s²
Explanation:
Let m₁ = 5 kg and m₂ = 10 kg. The figure is attached and free body diagrams of the objects are also attached.
Both objects (m₁ and m₂) have the same magnitude of acceleration(a). Let g be the acceleration due to gravity = 9.8 m/s². Hence:
T = m₁a (1)
m₂g - T = m₂a (2)
substituting T = m₁a in equation 2:
m₂g - m₁a = m₂a
m₂a + m₁a = m₂g
a(m₁ + m₂) = m₂g
a = m₂g / (m₁ + m₂)
a = (10 kg * 9.8 m/s²) / (10 kg + 5 kg) = 6.53 m/s²
Both objects have an acceleration of 6.53 m/s²
