An object weighs 2.6 N in air and 2.2N when completely immersed in water. Determine the relative density of the object (2mks

A concert loudspeaker suspended high off the ground emits 34 W of sound power. A small microphone with a 1.0 cm2 area is 44 m fr

rjkz [21]

Answer:

Part A

I = 1.4 mW/m²

Part B

β = 91.46 dB

Explanation:

Part A

Sound intensity is the power per unit area of sound waves in a direction perpendicular to that area. Sound intensity is also called acoustic intensity.

For a spherical sound wave, the sound intensity is given by;

$I = \frac{P}{A}$

$I = \frac{P}{4\pi r^{2}}$

Where;

P is the source of power in watts (W)

I is the intensity of the sound in watt per square meter (W/m2)

r is the distance r away

Given:

P = 34 W,

A = 1.0 cm²

r = 44 m

The sound intensity at the position of the microphone is calculated to be;

$I = \frac{34}{4\pi (44)^{2}}$

I = 0.0013975 W/m²

≈ I = 0.0014 W/m² = 1.4 × 10⁻³ W/m²

I = 1.4 mW/m²

The sound intensity at the position of the microphone is 1.4 mW/m².

Part B

Sound intensity level or acoustic intensity level is the level of the intensity of a sound relative to a reference value. It is a a logarithmic quantity. It is denoted by β and expressed in nepers, bels, or decibels.

Sound intensity level is calculated as;

β $= 10log_{10}\frac{I}{I_{0}}$ dB

Where,

β is the Sound intensity level in decibels (dB)

I is the sound intensity;

I₀ is the reference sound intensity;

By pluging-in, I₀ is 1.0 × 10⁻¹² W/m²

∴ β $= 10log_{10}\frac{1.4 * 10^{-3} W/m^{2}}{1.0 * 10^{-12} W/m^{2}}$

β $= 10log_{10} (1.4 * 10^{9})$

β = 91.46 dB

The sound intensity level at the position of the microphone is 91.46 dB.

4 0

2 years ago

n 38 g rifle bullet traveling at 410 m/s buries itself in a 4.2 kg pendulum hanging on a 2.8 m long string, which makes the pend

Kitty [74]

Answer:

68cm

Explanation:

You can solve this problem by using the momentum conservation and energy conservation. By using the conservation of the momentum you get

$p_f=p_i\\mv_1+Mv_2=(m+M)v$

m: mass of the bullet

M: mass of the pendulum

v1: velocity of the bullet = 410m/s

v2: velocity of the pendulum =0m/s

v: velocity of both bullet ad pendulum joint

By replacing you can find v:

$(0.038kg)(410m/s)+0=(0.038kg+4.2kg)v\\\\v=3.67\frac{m}{s}$

this value of v is used as the velocity of the total kinetic energy of the block of pendulum and bullet. This energy equals the potential energy for the maximum height reached by the block:

$E_{fp}=E_{ki}\\\\(m+M)gh=\frac{1}{2}mv^2$

g: 9.8/s^2

h: height

By doing h the subject of the equation and replacing you obtain:

$(0.038kg+4.2kg)(9.8m/s^2)h=\frac{1}{2}(0.038kg+4.2kg)(3.67m/s)^2\\\\h=0.68m$

hence, the heigth is 68cm

4 0

2 years ago

In models of magnetic and electric fields, why are field vectors depicted by arrows?

taurus [48]

The arrows in models of magnetic and electric fields show both their magnitude and direction.

In Physics, a vector refers to a quantity that has both magnitude and direction. Hence, a vector always points in a given direction. The direction in which the arrow points is the direction of the vector in space.

In models of magnetic and electric fields, field vectors depicted by arrows because they represent both their magnitude and direction. The length of the arrow shows magnitude.

Learn more: brainly.com/question/102477

7 0

1 year ago

Tortoise and the Hare

Tanya [424]

The answer would be D. The tortoise moved at a constant velocity throughout the race; the hare stopped to rest periodically.

Hope this helped. Good luck!

3 0

2 years ago

Read 2 more answers

Figure 3 shows a bicycle of mass 15 kg resting in a vertical position, with the front and back

Vinil7 [7]

Explanation:

There are three forces on the bicycle:

Reaction force Rp pushing up at P,

Reaction force Rq pushing up at Q,

Weight force mg pulling down at O.

There are four equations you can write: sum of the forces in the y direction, sum of the moments at P, sum of the moments at Q, and sum of the moments at O.

Sum of the forces in the y direction:

Rp + Rq − (15)(9.8) = 0

Rp + Rq − 147 = 0

Sum of the moments at P:

(15)(9.8)(0.30) − Rq(1) = 0

44.1 − Rq = 0

Sum of the moments at Q:

Rp(1) − (15)(9.8)(0.70) = 0

Rp − 102.9 = 0

Sum of the moments at O:

Rp(0.30) − Rq(0.70) = 0

0.3 Rp − 0.7 Rq = 0

Any combination of these equations will work.

3 0

2 years ago

An object weighs 2.6 N in air and 2.2N when completely immersed in water. Determine the relative density of the object (2mks​

An object weighs 2.6 N in air and 2.2N when completely immersed in water. Determine the relative density of the object (2mks