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Over [174]
3 years ago
15

A student is blowing over a straw 25 cm long, which is closed at one end. If the sound produced has a frequency of 350 Hz, what

is the speed of sound in the straw?
Physics
1 answer:
NARA [144]3 years ago
7 0

Answer:

If the end of the tube is uncovered such that the air at the end of the tube can freely vibrate when the sound wave reaches it, then the end is referred to as an open end. If both ends of the tube are uncovered or open, the musical instrument is said to contain an open-end air column.

Explanation: hope this helps:)

You might be interested in
A vertical spring stretches 4.0 cm when a 12-g object is hung from it. The object is replaced with a block of mass 28 g that osc
marin [14]

Answer:

Time period of the osculation will be 0.0671 sec  

Explanation:

It is given a vertical spring is stretched by 4 cm

So change in length of the spring x = 4 cm = 0.04 m

Mass which is hung from it m = 12 gram = 0.012 kg

Sprig force will be equal to weight of the mass

So kx=mg

k\times 0.04=0.012\times 9.8

k = 244.7 N/m

Now new mass is m = 28 gram = 0.028 kg

So time period with new mass will be

T=2\pi \sqrt{\frac{m}{k}}

=2\times 3.14 \sqrt{\frac{0.028}{244.7}}=0.0671sec

4 0
3 years ago
What is the mass moment of inertia of a 20kg sphere with a radius of 0.2m about a point on the sphere's perimeter
Kobotan [32]

Answer:

I = M R^2 is the moment of inertia about a point that is a distance R from the center of mass (uniform distributed mass).

The moment  of inertia about the center of a sphere is 2 / 5 M R^2.

By the parallel axis theorem the moment of inertia about a point on the rim of the sphere is  I = 2/5 M R^2 + M R^2 = 7/5 M R^2

I = 7/5 * 20 kg * .2^2 m = 1.12 kg m^2

7 0
2 years ago
What is the angle of reflection of the incident angle is 35°?<br> 35<br> 25°<br> 55<br> 90
Anestetic [448]

Answer:

i think its the third one or second one

4 0
3 years ago
What is the momentum of and 200 kg car traveling south at 22 m/sec?
motikmotik
The momentum of the car is 4.4x10^3 kg•m/sec
5 0
3 years ago
Three long parallel wires each carry 2.0-A currents in the same direction. The wires are oriented vertically, and they pass thro
blagie [28]

Answer:

21.2\times 10^{-6} T

Explanation:

i  = magnitude of current in each wire = 2.0 A

a  = length of the side of the square = 4 cm = 0.04 m

r  = length of the diagonal of the square = \sqrt{2} a = \sqrt{2} (0.04) = 0.057 m

B = magnitude of magnetic field by wires at A and C

B = \left ( \frac{\mu _{o}}{4\pi } \right )\left ( \frac{2i}{a} \right )

B = (10^{-7}) \left ( \frac{2(2)}{0.04} \right )

B = 10\times 10^{-6} T

B' = magnitude of magnetic field by wire at B

B' = \left ( \frac{\mu _{o}}{4\pi } \right )\left ( \frac{2i}{r} \right )

B' = (10^{-7}) \left ( \frac{2(2)}{0.057} \right )

B' = 7.02\times 10^{-6} T

Net magnitude of the magnetic field at D is given as

B_{net} = \sqrt{B^{2}+B^{2}} + B'

B_{net} = \sqrt{2} B + B'

B_{net} = \sqrt{2} (10\times 10^{-6}) + (7.02\times 10^{-6})

B_{net} = 21.2\times 10^{-6} T

8 0
3 years ago
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