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3 years ago

Large amplitude of sound vibrations will produce.....

Physics

1 answer:

user100 [1]3 years ago

8 0

Answer:

louder sound.

Explanation:

The amplitude of a sound wave determines its loudness or volume. A larger amplitude means a louder sound

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Determine the CM of a rod assuming its linear mass density λ (its mass per unit length) varies linearly from λ = λ0 at the left

Dahasolnce [82]

Answer:

$x_c= \dfrac{5}{9}L$

$I=\dfrac {7}{12}\lambda_ 0 L^3$

Explanation:

Here mass density of rod is varying so we have to use the concept of integration to find mass and location of center of mass.

At any distance x from point A mass density

$\lambda =\lambda_0+ \dfrac{2\lambda _o-\lambda _o}{L}x$

$\lambda =\lambda_0+ \dfrac{\lambda _o}{L}x$

Lets take element mass at distance x

dm =λ dx

mass moment of inertia

$dI=\lambda x^2dx$

So total moment of inertia

$I=\int_{0}^{L}\lambda x^2dx$

By putting the values

$I=\int_{0}^{L}\lambda_ ox+ \dfrac{\lambda _o}{L}x^3 dx$

By integrating above we can find that

$I=\dfrac {7}{12}\lambda_ 0 L^3$

Now to find location of center mass

$x_c = \dfrac{\int xdm}{dm}$

$x_c = \dfrac{\int_{0}^{L} \lambda_ 0(1+\dfrac{x}{L})xdx}{\int_{0}^{L} \lambda_0(1+\dfrac{x}{L})}$

Now by integrating the above

$x_c=\dfrac{\dfrac{L^2}{2}+\dfrac{L^3}{3L}}{L+\dfrac{L^2}{2L}}$

$x_c= \dfrac{5}{9}L$

So mass moment of inertia $I=\dfrac {7}{12}\lambda_ 0 L^3$ and location of center of mass $x_c= \dfrac{5}{9}L$

8 0

3 years ago

The amount of matter contained in a given volume for a substance is called

Valentin [98]

This is called density.

5 0

3 years ago

you see a lightning bolt in the sky. You hear a clap of thunder 3 seconds latter. The speed of 330 m/s. How far away was the lig

Naddika [18.5K]

The i got answer is 990m

8 0

2 years ago

2. A construction consultant may be responsible for

Brrunno [24]

This is a tricky one but on my part I'd have to say depending on the contract A,B,C.

3 0

3 years ago

If the earth's magnetic field has strength 0.50 gauss and makes an angle of 20.0 degrees with the garage floor, calculate the ch

lys-0071 [83]

Answer:

ΔΦ = -3.39*10^-6

Explanation:

Given:-

- The given magnetic field strength B = 0.50 gauss

- The angle between earth magnetic field and garage floor ∅ = 20 °

- The loop is rotated by 90 degree.

- The radius of the coil r = 19 cm

Find:

calculate the change in the magnetic flux δφb, in wb, through one of the loops of the coil during the rotation.

Solution:

- The change on flux ΔΦ occurs due to change in angle θ of earth's magnetic field B and the normal to circular coil.

- The strength of magnetic field B and the are of the loop A remains constant. So we have:

Φ = B*A*cos(θ)

ΔΦ = B*A*( cos(θ_1) - cos(θ_2) )

- The initial angle θ_1 between the normal to the coil and B was:

θ_1 = 90° - ∅

θ_1 = 90° - 20° = 70°

The angle θ_2 after rotation between the normal to the coil and B was:

θ_2 = ∅

θ_2 = 20°

- Hence, the change in flux can be calculated:

ΔΦ = 0.5*10^-4*π*0.19*( cos(70) - cos(20) )

ΔΦ = -3.39*10^-6

8 0

3 years ago