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2 years ago

In which position would an object have a greater amount of potential energy

Physics

2 answers:

valentinak56 [21]2 years ago

7 0

Answer:

At the maximum height

Explanation:

The gravitational potential energy of an object is the energy possessed by the object due to its position in the gravitational field.

Mathematically, it is given by:

$PE=mgh$

where

m is the mass of the object

g is the acceleration due to gravity

h is the height of the object relative to the ground

From the formula, we see that the potential energy is directly proportional to the height of the object above the ground: therefore, the larger the value of h, the larger the potential energy of the object.

Therefore, the object has the maximum potential energy when it is at the highest position above the ground.

Ludmilka [50]2 years ago

6 0

Answer: hanging 20 m from the ground

Explanation:

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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.

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Explanation:

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u is the initial speed

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