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3 years ago

By what factor must the amplitude of a sound wave be decreased in order to decrease the intensity by a factor of 3?

Physics

1 answer:

s344n2d4d5 [400]3 years ago

4 0

Answer:

Amplitude is decreased by a factor of $\sqrt3$ if intensity is decreased by a factor of 3.

Explanation:

Intensity of a sound wave is directly proportional to the square of its amplitude.

Therefore, if intensity is $I$ and amplitude is $A$ , then

$I=kA^2$ , where, $k$ is constant of proportionality.

Now, if intensity of sound wave is decreased by a factor of 3. So,

New intensity is, $I_{new}=\frac{I}{3}$

$I_{new}=kA_{new}^2\\\frac{I}{3}=kA_{new}^2$

Plug in $kA^2$ for $I$ . This gives,

$\frac{kA^2}{3}=kA_{new}^2\\A_{new}^2=\frac{A^2}{3}\\A_{new}=\sqrt{\frac{A^2}{3}}=\frac{A}{\sqrt{3}}$

Therefore, amplitude is decreased by a factor of $\sqrt3$ .

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A golf ball stays on the tee until the golf club hits it. Which of the following principles best describes why this occurs? Grou

givi [52]

Answer:

Newton's First Law of Motion.

Explanation:

Newton's first law of motion states that an object continues to stay in its state of rest, or of uniform motion, until acted upon by an external force.

So in the case of the golf ball here, the ball stays in its state of rest, on the tee, until the golf club hits it, i.e. , applies an external force on it.

Hence we can say that Newton's First Law of Motion is the principle which is most suitable for explaining this phenomenon.

7 0

4 years ago

as you go above earths surface, the acceleration due to gravity will decrease. find the height, i meters, above the earths surfa

aleksklad [387]

Answer:

1.23×10⁸ m

Explanation:

Acceleration due to gravity is:

a = GM / r²

where G is the universal gravitational constant,

M is the mass of the planet,

and r is the distance from the center of the planet to the object.

When the object is on the surface of the Earth, a = g and r = R.

g = GM / R²

When the object is at height i above the surface, a = 1/410 g and r = i + R.

1/410 g = GM / (i + R)²

Divide the first equation by the second:

g / (1/410 g) = (GM / R²) / (GM / (i + R)²)

410 = (i + R)² / R²

410 R² = (i + R)²

410 R² = i² + 2iR + R²

0 = i² + 2iR − 409R²

Solve with quadratic formula:

i = [ -2R ± √((2R)² − 4(1)(-409R²)) ] / 2(1)

i = [ -2R ± √(1640R²) ] / 2

i = (-2R ± 2R√410) / 2

i = -R ± R√410

i = (-1 ± √410) R

Since i > 0:

i = (-1 + √410) R

R = 6.37×10⁶ m:

i ≈ 1.23×10⁸ m

8 0

3 years ago

Please help me with this and if I have good answer u will be marked as brilliant !!!!

AysviL [449]

Number one- numbers of items sold. Number two- Thursday and Friday. Number three- 1,200. Number four-150

3 0

4 years ago

Use the conservation of energy to explain the speeds at different places in the diagram of the roller coaster.

damaskus [11]

When you are going down you pick up more speed

4 0

3 years ago

An electron passes through a point 2.83 cm 2.83 cm from a long straight wire as it moves at 35.5 % 35.5% of the speed of light p

igor_vitrenko [27]

Answer:

The magnitude of electron acceleration is $2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

Explanation:

Given:

Distance from the wire to the field point $r = 2.83 \times 10^{-2}$ m

Speed of electron $v = 35.5 \%c$

Current $I = 17.7$ A

For finding the acceleration,

First find the magnetic field due to wire,

$B = \frac{\mu _{o}I }{2\pi r }$

Where $\mu_{o} = 4\pi \times 10^{-7}$

$B = \frac{4\pi \times 10^{-7} \times 17.7 }{2\pi (2.83 \times 10^{-2} ) }$

$B = 12.50 \times 10^{-5}$ T

The magnetic force exerted on the electron passing through straight wire,

$F = qvB$

$F = 1.6 \times 10^{-19} \times 0.355 \times 3 \times 10^{8} \times 12.50 \times 10^{-5}$

$F = 21.3 \times 10^{-16}$ N

From the newton's second law

$F = ma$

Where $m =$ mass of electron $= 9.1 \times 10^{-31}$ kg

So acceleration is given by,

$a = \frac{F}{m}$

$a = \frac{21.3 \times 10^{-16} }{9.1 \times 10^{-31} }$

$a = 2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

Therefore, the magnitude of electron acceleration is $2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

7 0

3 years ago