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frosja888 [35]
3 years ago
8

The loudness of a sound is determined by the __________, or height, of the sound wave.

Physics
2 answers:
Fittoniya [83]3 years ago
6 0

c

just took the quiz

balandron [24]3 years ago
5 0

C.    amplitude....................


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A wheel that was initially spinning is accelerated at a constant angular acceleration of 5.0 rad/s^2. After 8.0 s, the wheel is
notka56 [123]

Answer:

a)  Initial angular speed = 30 rad/s

b) Final angular speed = 70 rad/s        

Explanation:

a) We have equation of motion s = ut + 0.5at²

    Here s = 400 radians

              t = 8 s

              a = 5 rad/s²

    Substituting

             400 = u x 8 + 0.5 x 5 x 8²

              u = 30 rad/s

   Initial angular speed = 30 rad/s

b) We have equation of motion v = u + at

     Here u = 30 rad/s

               t = 8 s

              a = 5 rad/s²  

    Substituting

             v = 30 + 5 x 8 = 70 rad/s    

   Final angular speed = 70 rad/s        

8 0
3 years ago
What happens when stress builds up at faults?
Harlamova29_29 [7]

Answer:

The answer is D because when the faults move that is the tectonic plates moving. So earth quakes will be forming when the fault moves.

Explanation:

3 0
3 years ago
Read 2 more answers
You lift a 10-N physics book up in the air a distance of 1.0 m, at a constant velocity
Anarel [89]

The book is lifted upward, but gravity points down, so the work done by gravity must be negative (so you can eliminate options 1 and 3).

The force exerted on the book by gravity has magnitude

<em>F</em> = <em>mg</em> = (10 N) (9.80 m/s^2) = 9.8 N ≈ 10 N

You raise the book 1.0 m in the opposite direction, so the work done is

<em>W</em> = (10 N) (-1.0 m) = -10 J

5 0
3 years ago
You have a pendulum clock made from a uniform rod of mass M and length L pivoting around one end of the rod. Its frequency is 1
drek231 [11]

The new oscillation frequency of the pendulum clock is 1.14 rad/s.

     

The given parameters;

  • <em>Mass of the pendulum, = M </em>
  • <em>Length of the pendulum, = L</em>
  • <em>Initial angular speed, </em>\omega _i<em> = 1 rad/s</em>

The moment of inertia of the rod about the end is given as;

I_i = \frac{1}{3} ML^2

The moment of inertia of the rod between the middle and the end is calculated as;

I_f = \int\limits^L_{L/2} {r^2\frac{M}{L} } \, dr = \frac{M}{3L} [r^3]^L_{L/2} =  \frac{M}{3L} [L^3 - \frac{L^3}{8} ] = \frac{M}{3L} [\frac{7L^3}{8} ]= \frac{7ML^2}{24}

Apply the principle of conservation of angular momentum as shown below;

I _i \omega _i = I _f \omega _f\\\\\frac{ML^2}{3} (1 \ rad/s)= \frac{7ML^2}{24} \times \omega _f\\\\\frac{24 \times ML^2}{3 \times 7 ML^2} (1 \ rad/s)= \omega _f\\\\1.14 \ rad/s = \omega _f

Thus, the new oscillation frequency of the pendulum clock is 1.14 rad/s.

Learn more about moment of inertia of uniform rod here: brainly.com/question/15648129

3 0
3 years ago
Please help me , thank you
mash [69]

Answer:

Since the reading wasn't specified, it would be most likely A

Explanation:

A is the most similar to a protoplanetary disk, so it'd be A most likely

5 0
3 years ago
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