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Leokris [45]
4 years ago
9

A 60 cm diameter wheel accelerates from rest at a rate of 7 rad/s2. What is the tangential acceleration (in m/s2) of a point on

the edge of the wheel?
Physics
1 answer:
wolverine [178]4 years ago
3 0

Answer:

2.1 m/s²

Explanation:

By definition, the angular acceleration, is equal to the rate of change of the angular velocity, ω:

α = Δω / Δt (1)

By definition of the angular velocity, we can express the linear velocity, v, as follows:

v = ω*r⇒ Δv = Δω*r (2)

Replacing Δω, from (1) in (2), we get:

Δv = α*Δt*r⇒ Δv/Δt = α*r (3)

By definition of linear acceleration, we can write the following expression;

a = α*r

For a point on the edge of the wheel, the linear acceleration is tangent to the rim, and is equal to the product of the angular acceleration times the distance to the center, which for a point on the edge of the wheel, is just the radius:

⇒ a = 7 rad/sec²*0.3m = 2.1 m/s²

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Answer:

It will cause kinetic energy to increase.

Explanation:

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Read 2 more answers
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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