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Firlakuza [10]
3 years ago
14

Which of the following conditions would always result in a net force of 0N?

Physics
2 answers:
9966 [12]3 years ago
7 0

A condition which would always result in a net force of zero (0) Newton is: A. Two equal but opposite forces act on an object.

Newton's Second Law of Motion states that the acceleration of an object is inversely proportional to its mass and directly proportional to the net force acting on the physical object.

Mathematically, Newton's Second Law of Motion is given by this formula;

Acceleration = \frac{Net \; force}{Mass}

According to Newton's Second Law of Motion, the acceleration of a physical object occurs when there is a net force on an object while a net force (sum of all the forces) zero (0) causes no acceleration.

In Science, an object is said to be at equilibrium when two equal but opposite forces are acting on it and this would always result in a net force of zero (0) Newton.

<u>For example:</u>

0 N = 5N + (-5N)\\\\0N = 5N - 5N

Read more here: brainly.com/question/24029674

grigory [225]3 years ago
4 0

Answer:

An object is at rest

Explanation:

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3 years ago
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:

<u>Part A</u>

I = 1.4 mW/m²  

<u>Part B</u>

β = 91.46 dB

Explanation:

<u>Part A</u>

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 = \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.

<u>Part B</u>

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.                

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