Atoms involved in blank covalent bonds do not share their electrons equally

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

3 years ago

Help me please !!!!!

bezimeni [28]

Answer:

confounding cause they had exposure to many programmes

8 0

3 years ago

Read 2 more answers

It takes 52,000 Joules to heat a cup of coffee to boiling from room temperature. How long a piece of 20 cm wide Aluminum foil wo

vovikov84 [41]

Answer:

L = 1.11 x $10^{6}$ m, is the length of piece of 20 cm wide Aluminum foil to make capacitor large enough to hold 52000 J of energy.

Explanation:

Solution:

Data Given:

Heat Energy = 52000 J

Dielectric Constant of the plastic Bag = 3.7 = K

Thickness = 2.6 x $10^{5}$ m =d

V = 610 volts

A = width x Length

width = 20 cm = 20 x $10^{-2}$ m

Length = ?

So,

we know that,

U = 1/2 C Δ $v^{2}$

U = 52000 J

C = ?

V = 610 volts'

So,

U = 1/2 C Δ $v^{2}$

52000 J = (0.5) x (C) x ( $610^{2}$ )

C = 0.28 F

And we also know that,

C = $\frac{K*E*A}{d}$

E = 8.85 x $10 ^{-12}$

K = 3.7

A = 0.20 x L

d = 2.6 x $10^{5}$ m

Plugging in the values into the formula, we get:

0.28 = $\frac{3.7 * 8.85 .10^{-12} * (0.20 . L) }{2.6 . 10^{5} }$

Solving for L, we get:

L = 1.11 x $10^{6}$ m,

is the length of piece of 20 cm wide Aluminum foil to make capacitor large enough to hold 52000 J of energy.

7 0

3 years ago

a force of 17 acts on a block of mass 4kg the forces of friction opposing the motion is 5n . the acceleration of the block is

Nady [450]

Fk = μK N

N = m a

N = 4 × a

N = 4a

Fk = μK N

17 = 5 × 4a

17 = 20 a

a = ¹⁷/₂₀ = 0.85 m/s²

8 0

2 years ago

Which factors affect heat transfer between a warm and a cool substance?

andrey2020 [161]

Answer: B

Explanation:

It's not the time it took to heat the substance, so that rules out A and C.

This means that we only have to choose between

B. the area of contact

D. the area of the substances

(since everything else in each of those answers are the same)

Area of contact matters more (e.g. an object with greater surface area is exposed to the air more, will lose/gain heat quicker than an object with less surface area).

5 0

3 years ago