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svetoff [14.1K]
3 years ago
9

A pure sound wave, generated by a tuning fork, is considered a periodic wave. Which statement is true for this tuning fork sound

wave?
A. The wave pattern repeats itself after every two waves.
B. Every wave has the same wave pattern.
C. The amplitude of the waves increases after every subsequent wave.
D. No two waves have the same wave pattern.
Physics
2 answers:
soldier1979 [14.2K]3 years ago
4 0

Answer : Every wave has the same wave pattern.

Explanation : A pure sound wave has a single frequency and its generated by a tuning fork . The sound wave is a  simple periodic wave.

When an object such as a guitar push and pull on the around it. When it pushes on the air, then the pressure increases and when it pull on the air,  then the pressure decreases and sound waves are formed.

Hence, sound wave has the same wave pattern.  

ch4aika [34]3 years ago
4 0

Answer:

Option B

Explanation:

A periodic wave has repeating pattern depending on which the wavelength and frequency of the wave can be determined. A pure sound wave which is generated by a tuning fork, considered as a periodic wave has the same wave pattern for every wave.

Thus, option B is correct.

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At the surface, atmospheric pressure is 1.013 × 10^5 Pa. People can normally snorkel down to a depth of roughly one meter. What
natulia [17]

Answer:

1.01 × 10⁵ Pa  

Explanation:

At the surface, atmospheric pressure is 1.013 × 10⁵ Pa.

We need to find the total pressure on the air in the lungs of a person to a depth of 1 meter.

Pressure at a depth is given by :

P=\rho gh

Where

\rho is the density of air, \rho=1.225\ kg/m^3

So,

P=1.225\times 9.8\times 1\\\\=12\ Pa

Total pressure, P = Atmospheric pressure + 12 Pa

= 1.013 × 10⁵ Pa + 12 Pa

= 1.01 × 10⁵ Pa

Hence, the total pressure is 1.01 × 10⁵ Pa.

5 0
3 years ago
A mass of 5kg starts from rest and pulls down vertically on a string wound around a disk-shaped, massive pulley. The mass of the
Paha777 [63]

Answer:

c. V = 2 m/s

Explanation:

Using the conservation of energy:

E_i =E_f

so:

Mgh = \frac{1}{2}IW^2 +\frac{1}{2}MV^2

where M is the mass, g the gravity, h the altitude, I the moment of inertia of the pulley, W the angular velocity of the pulley and V the velocity of the mass.

Also we know that:

V = WR

Where R is the radius of the disk, so:

W = V/R

Also, the moment of inertia of the disk is equal to:

I = \frac{1}{2}MR^2

I = \frac{1}{2}(5kg)(2m)^2

I = 10 kg*m^2

so, we can write the initial equation as:

Mgh = \frac{1}{2}IV^2/R^2 +\frac{1}{2}MV^2

Replacing the data:

(5kg)(9.8)(0.3m) = \frac{1}{2}(10)V^2/(2)^2 +\frac{1}{2}(5kg)V^2

solving for V:

(5kg)(9.8)(0.3m) = V^2(\frac{1}{2}(10)1/4 +\frac{1}{2}(5kg))

V = 2 m/s

8 0
3 years ago
A hollow cylinder that is rolling without slipping is given a velocity of 5.0 m/s and rolls up an incline to a vertical height o
inysia [295]

Answer:

The hollow cylinder rolled up the inclined plane by 1.91 m

Explanation:

From the principle of conservation of mechanical energy, total kinetic energy = total potential energy

M.E_T = \frac{1}{2}mv^2 + \frac{1}{2} I \omega^2 + mgh

The total energy at the bottom of the inclined plane = total energy at the top of the inclined plane.

\frac{1}{2}mv_i^2 + \frac{1}{2} I \omega_i^2 + mg(0) =  \frac{1}{2}mv_f^2 + \frac{1}{2} I \omega_f^2 + mgh

moment of inertia, I, of a hollow cylinder = ¹/₂mr²

substitute for I in the equation above;

\frac{1}{2}mv_i^2 + \frac{1}{2} (\frac{1}{2}mr^2  \omega_i^2) =  \frac{1}{2}mv_f^2 + \frac{1}{2} (\frac{1}{2}mr^2  \omega_f^2) + mgh\\\\ but \ v = r \omega\\\\\frac{1}{2}mv_i^2 + \frac{1}{2} (\frac{1}{2}m v_i^2  ) =  \frac{1}{2}mv_f^2 + \frac{1}{2} (\frac{1}{2}m v_f^2) + mgh\\\\\frac{1}{2}mv_i^2 +\frac{1}{4}mv_i^2 = \frac{1}{2}mv_f^2 +\frac{1}{4}mv_f^2 +mgh\\\\\frac{3}{4}mv_i^2 = \frac{3}{4}mv_f^2 +mgh\\\\mgh = \frac{3}{4}mv_i^2 -  \frac{3}{4}mv_f^2\\\\gh = \frac{3}{4}v_i^2 -  \frac{3}{4}v_f^2\\\\

h = \frac{3}{4g}(v_1^2 -v_f^2)

given;

v₁ = 5.0 m/s

vf = 0

g = 9.8 m/s²

h = \frac{3}{4g}(v_1^2 -v_f^2) =\frac{3}{4*9.8}(5^2 -0) = 1.91 \ m

Therefore, the hollow cylinder rolled up the inclined plane by 1.91 m

5 0
3 years ago
A force off 700 newtons is applied to a 600 kg bowling ball. What is the acceleration of the bowling ball once the force is appl
Readme [11.4K]

Answer:

<h2>1.17 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

f is the force

m is the mass

From the question we have

a =  \frac{700}{600}  =  \frac{7}{6}  \\  = 1.1666666...

We have the final answer as

<h3>1.17 m/s²</h3>

Hope this helps you

6 0
3 years ago
One billiard ball is shot east at 2.00 m/s. A second, identical billiard ball is shot west at 1.00 m/s. The balls have a glancin
dimulka [17.4K]

Answer:

Velocity is 1.73 m/s along 54.65° south of east.

Explanation:

Let unknown velocity be v, mass of billiard ball be m and east direction be positive x axis.

Here momentum is conserved.

Initial momentum = Final momentum

Initial momentum = m x 2i + m x (-1)i = m i

Final momentum = m x v + m x 1.41 j = mv + 1.41 m j

Comparing

mi = mv + 1.41 m j

v = i - 1.41 j

Magnitude of velocity

      v=\sqrt{1^2+(-1.41)^2}=1.73m/s        

Direction,  

       \theta =tan^{-1}\left ( \frac{-1.41}{1}\right )=-54.65^0             

Velocity is 1.73 m/s along 54.65° south of east.

5 0
3 years ago
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