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2 years ago

Calculate the amount of air in a room 6m long, 5m wide and 3mm high.

Physics

2 answers:

sattari [20]2 years ago

6 0

Answer:

0.09kg of air

Explanation:

The dimensions of the room are given

change the height to meters by dividing it by thousand.

For the volume multiplying the length,width and height (all should be in the same unit most suitable being meters).

Volume refers to the amount of space inside a box or a object.

The amount of air is equal to the volume.

AlexFokin [52]2 years ago

3 0

Answer:

90 m^3

Explanation:

Volume of the room:

6 m * 5 m * 3 m = 90 m^3 <=====( I changed 3mm to 3 m)

if 3mm is not a typo mistake

volume becomes ( 3 mm = .003 m)

6 m * 5 m * .003 m = .09 m^3 ( though unlikely )

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a 3 kg piece of putty that is moving with a velocity of 10 m/s collides and sticks to an 8 kg bowling ball that was at rest. wha

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The final velocity is 2.7 m/s

Explanation:

We can solve this problem by using the principle of conservation of momentum: in fact, in absence of external forces, the total momentum of the system must be conserved before and after the collision.

Therefore we can write:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v$

where:

$m_1 = 3 kg$ is the mass of the putty

$u_1 = 10 m/s$ is the initial velocity of the putty (we take its direction as positive direction)

$m_2 = 8 kg$ is the mass of the ball

$u_2 = 0 m/s$ is the initial velocity of the ball (at rest)

$v$ is the final combined velocity of the two putty+ball

Re-arranging the equation and substituting the values, we find the final combined velocity:

$v=\frac{m_1 u_1 + m_2 u_2}{m_1+m_2}=\frac{(3)(10)+0}{3+8}=2.7 m/s$

And the positive sign indicates their final direction is the same as the initial direction of the putty.

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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

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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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Calculate the amount of air in a room 6m long, 5m wide and 3mm high.​

Calculate the amount of air in a room 6m long, 5m wide and 3mm high.