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3 years ago

A sound wave generated by a musical note has the characteristics presented

Physics

2 answers:

Anuta_ua [19.1K]3 years ago

6 0

Answer:

Explanation:subtract all of those by the all of the other numbers and that’s the answer i think that’s the way I learned it

dlinn [17]3 years ago

4 0

Answer: its B 6.3

Explanation:

AP3X

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How can you increase the intensity of sound waves

VladimirAG [237]

Answer

The intensity of a sound wave depends on the pressure of the wave,density of the medium and speed of sound in the medium. Higher density and higher sound speed both give a lower intensity. and may be it is because that sound wave is more characterize by wavelength than frequency..explanation

Explanation:

As decibel levels get higher, sound waves have greater intensity and sounds are louder. For every 10-decibel increase in the intensity of sound, loudness is 10 times greater. Intensity of sound results from two factors: the amplitude of the sound waves and how far they have traveled from the source of the sound.

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3 years ago

Why the effect of gravitational force is more in liquid than in solid?

LekaFEV [45]

Gravitational force depends only on mass and distance, not on the state of matter.
The forces of attraction between molecules in matter are electromagnetic in nature, not gravitational.
These attractive forces are stronger in a solid than in a liquid than in a gas.
Gravitational forces between molecules is completely negligible compared to the em forces.

So, key answer is inter-molecular forces of solids is stronger than liquids.

4 0

3 years ago

Read 2 more answers

What is the average acceleration of the particle between 0 seconds and 4 seconds? A. 0 meters/second2 B. 0.04 meters/second2 C.

lisov135 [29]

Answer

D. 0.25 meters/second2

Explanation

The average acceleration is the ratio of change in velocity to the change in time of travel.Taking in this case that the change of velocity is a unit, then Average acceleration is given by;

Aacc=Vf-Vi/Tf-Ti

where Vf=final velocity,Vi=initial velocity' Tf=final time, Ti=initial time

Vf-Vi=1m/s

Tf-Ti=4-0=4seconds

Avacc=1/4=0.25m/s2

6 0

3 years ago

A loaded 375 kg toboggan is traveling on smooth horizontal snow at 4.50 m/s when it suddenly comes to a rough region. The region

zmey [24]

Answer:

a) The average friction force exerted on the toboggan is 653.125 newtons, b) The rough region reduced the kinetic energy of the toboggan in 92.889 %, c) The speed of the toboggan is reduced in 73.333 %.

Explanation:

a) Given the existence of non-conservative forces (friction between toboggan and ground), the motion must be modelled by means of the Principle of Energy Conservation and the Work-Energy Theorem, since toboggan decrease its speed (associated with due to the action of friction. Changes in gravitational potential energy can be neglected due to the inclination of the ground. Then:

$K_{1} = K_{2} + W_{f}$

Where:

$K_{1}$ , $K_{2}$ are the initial and final translational kinetic energies of the tobbogan, measured in joules.

$W_{f}$ - Dissipated work due to friction, measured in joules.

By applying definitions of translation kinetic energy and work, the expression described above is now expanded and simplified:

$f\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{1}^{2}-v_{2}^{2})$

Where:

$f$ - Friction force, measured in newtons.

$\Delta s$ - Distance travelled by the toboggan in the rough region, measured in meters.

$m$ - Mass of the toboggan, measured in kilograms.

$v_{1}$ , $v_{2}$ - Initial and final speed of the toboggan, measured in meters per second.

The friction force is cleared:

$f = \frac{m\cdot (v_{1}^{2}-v_{2}^{2})}{2\cdot \Delta s}$

If $m = 375\,kg$ , $v_{1} = 4.50\,\frac{m}{s}$ , $v_{2} = 1.20\,\frac{m}{s}$ and $\Delta s = 5.40 \,m$ , then:

$f = \frac{(375\,kg)\cdot \left[\left(4.50\,\frac{m}{s} \right)^{2}-\left(1.20\,\frac{m}{s}\right)^{2}\right]}{2\cdot (5.40\,m)}$

$f = 653.125\,N$

The average friction force exerted on the toboggan is 653.125 newtons.

b) The percentage lost by the kinetic energy of the tobbogan due to friction is given by the following expression, which is expanded and simplified afterwards:

$\% K_{loss} = \frac{K_{1}-K_{2}}{K_{1}}\times 100\,\%$