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Brums [2.3K]
3 years ago
6

An auditorium measures 35.0 m x 30.0 m x 5.0 m. The density of air is 1.20 kg/m^3. (a) What is the volume of the room in cubic f

eet? (b) What is the weight of air in the room in pounds?
Physics
1 answer:
Ksenya-84 [330]3 years ago
5 0

Answer:

(a) 1852259 ft^3 (b) 489085.47 pound

Explanation:

We have given auditorium measures 35 m×30 m×5 m

We know that 1 meter = 3.28 feet

So the measure of auditorium = 35×3.28 feet ×30×3.28 feet× 5×3.28 feet

(a) So the volume of the auditorium =35\times 3.28\times 30\times 3.28\times 5\times 3.28=185259.648ft^3

Density is given as d=1.20kg/m^3  

(b) weight of air  = volume × density =185259.648\times 1.2=222311.577kg

We know that 1 kg = 2.20 pound

So 222311.577 kg =222311.577×2.20=489085.47 pound

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You drop a steel ball bearing, with a radius of 2.40 mm, into a beaker of honey. Note that honey has a viscosity of 6.00 Pa/s an
Stells [14]

Answer:

The “terminal speed” of the ball bearing is 5.609 m/s

Explanation:

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Viscosity of honey η = 6.0 Pa/s

\text { Viscosity has Density } \sigma=1360 \mathrm{kg} / \mathrm{m}^{3}

\text { Steel has a density } \rho=7800 \mathrm{kg} / \mathrm{m}^{3}

\left.\mathrm{g}=9.8 \mathrm{m} / \mathrm{s}^{2} \text { (g is referred to as the acceleration of gravity. Its value is } 9.8 \mathrm{m} / \mathrm{s}^{2} \text { on Earth }\right)

While calculating the terminal speed in liquids where density is high the stokes law is used for viscous force and buoyant force is taken into consideration for effective weight of the object. So the expression for terminal speed (Vt)

V_{t}=\frac{2 \mathrm{R}^{2}(\rho-\sigma) \mathrm{g}}{9 \eta}

Substitute the given values to find "terminal speed"

\mathrm{V}_{\mathrm{t}}=\frac{2 \times 0.0024^{2}(7800-1360) 9.8}{9 \times 6}

\mathrm{V}_{\mathrm{t}}=\frac{0.0048 \times 6440 \times 9.8}{54}

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\mathrm{V}_{\mathrm{t}}=5.609 \mathrm{m} / \mathrm{s}

The “terminal speed” of the ball bearing is 5.609 m/s

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Productivity increases when A. inputs increase while outputs remain the same. B. inputs decrease while outputs remain the same.
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The most precise measurement of the speed of light in vacuum is 299,792,458 m/s. Which of the following values would you use to
Luda [366]

Answer:

v = 3.00 x 10⁸ m/s

Explanation:

given,

speed of light in vacuum = 299,792,458 m/s

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