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Varvara68 [4.7K]
2 years ago
11

A trombone has a variable length. When a musician blows into the mouthpiece and causes air in the tube of the horn to vibrate, t

he waves set up by the vibrations reflect back and forth in the horn to create standing waves. As the length of horn is made shorter, what happens to the frequency?
Physics
1 answer:
Lilit [14]2 years ago
6 0

Answer:

The frequency increases.

Explanation:

When the Musician draws the slide in the length of the horn gets shorter, which causes a decrease in the wavelength. A decrease in the wave length results in an increase in frequency.

Note:

The diameter of the horn has an effect on frequency, so a wider horn is effectively a long horn - open end correction ( distance between the the antinode and the open end of a pipe).

Frequency also depends on how hard the musician blows the trombone. The musician can change the frequency with the lip pressure being applied.

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An unlabeled hierarchical diagram of various astronomical bodies is shown below. The labels A, B, C, and D can be used to repres
Paha777 [63]
A) The biggest astronomical object is the Universe, which contains billions of galaxies among which there is the Milky Way.
The Milky Way contains thousands of planetary systems, among which the Solar System.
The Solar System contains many <span>planets <span>(but only one star, the Sun)</span>,</span> among which there is Earth.
Therefore you can label:
A = Universe, B = Milky Way, C = Solar system, D = Earth

b) Given what we said before, you could label D also any other planet in the Solar System, therefore you can choose among Mercury, Venus, Mars, Jupiter, Saturn, Uranus, and Neptune.
8 0
3 years ago
ball of mass 0.4 kg is attached to the end of a light stringand whirled in a vertical circle of radius R = 2.9 m abouta fixed po
Levart [38]

Answer:

6.046N

Explanation:

The net force exerted on the mass is the sum of tension force and the external force of gravity.

F_n_e_t=F_g+F_t

F_t is the tension force.F_g=9.8N/kg is the force of gravity.

F_n_e_t=ma_c=mv^2/r\\

where r is the rope's radius from the fixed point.

From the net force equation above:

F_t=F_n_e_t-F_g\\=mv^2/r-mg\\=0.4\times(8.5^2/2.9)-0.4\times9.8\\=6.046N

Hence the tension force is 6.046N

8 0
3 years ago
When force is applied to a breaker bar the torque can be calculated by multiplying the length of the lever by the?
Nimfa-mama [501]

When a force applied to a breaker bar the torque can be calculated by multiplying the<u> length of the lever</u> by the tangential component of force on the lever.

<h3>What is torque?</h3>

Torque is the <u>rotating equivalent</u> of force in physics and mechanics. Depending on the subject of study, it is also known as the moment, moment of force, rotating force, or turning effect. It illustrates how a force can cause a change in the body's rotational motion.

Torque is given by the formula :

                          α = r x F ( bold letters represent vector quantities)

The S.I. unit for torque is :  N - m ( Newton - meter)

<h3>How do we define 1 N-m of torque?</h3>

The newton-metre is a torque unit (also known as a moment) in the SI system. The torque produced by a one newton force applied <u>perpendicularly to the end of a one metre long</u> moment arm is known as a newton-metre.

To learn more about torque:

brainly.com/question/14970645

#SPJ4

5 0
1 year ago
3. Identify the structures of the skin by the labels A through G below
ioda
I need help to I have a big test and I need help so bad
4 0
3 years ago
Consider a 2.54-cm-diameter power line for which the potential difference from the ground, 19.6 m below, to the power line is 11
tiny-mole [99]

Answer:

The line charge density is 1.59\times10^{-4}\ C/m

Explanation:

Given that,

Diameter = 2.54 cm

Distance = 19.6 m

Potential difference = 115 kV

We need to calculate the line charge density

Using formula of potential difference

V=EA

V=\dfrac{\lambda}{2\pi\epsilon_{0}r}\times\pi r^2

\lambda=\dfrac{V\times2\epsilon_{0}}{r}

Where, r = radius

V = potential difference

Put the value into the formula

\lambda=\dfrac{115\times10^{3}\times2\times8.8\times10^{-12}}{1.27\times10^{-2}}

\lambda=1.59\times10^{-4}\ C/m

Hence, The line charge density is 1.59\times10^{-4}\ C/m

4 0
3 years ago
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