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Colt1911 [192]
2 years ago
13

Clouds can weigh thousands of pounds due to their liquid water content. Often this content is measured in grams per cubic meter

(g/m3). Assume that a cumulus cloud occupies a volume of one cubic kilometer, and its liquid water content is 0.2 g/m3. (a) What is the volume of this cloud in cubic miles? (b) How much does the water in the cloud weigh in pounds?
Physics
1 answer:
-Dominant- [34]2 years ago
8 0

Answer:

a) 0.2399 mi³

b) 440.8 × 10³ Pounds

Explanation:

Given:

Volume of cumulus cloud, V = 1 km³

Liquid water content = 0.2 g/m³

Now,

a) 1 km = \frac{\textup{1 miles}}{\textup{1.6093}}

thus,

1 km³ = (\frac{\textup{1 miles}}{\textup{1.6093}})^3

1 km³ =  0.2399 mi³

Hence, volume of cloud in cubic miles is 0.2399 mi³

b)

Liquid water content = 0.2 g/m³

Now,

1 Km = 1000 m

thus,

1 km³ = 1000³ m³

Therefore,

Liquid water content in 1 Km³ of cloud = 0.2 g/m³ × 1000³ m³

= 200 × 10⁶ gram

or

= 200 × 10³ Kg

also,

1 kilogram =  2.204 pounds

Therefore,

200 × 10³ Kg = 200 × 10³ × 2.204 pounds = 440.8 × 10³ Pounds

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Scorpion4ik [409]

If gravitational effects from other objects are negligible, the speed of the rock at a very great distance from the planet will approach a value of \sqrt{3} v_{e}

<u>Explanation:</u>

To express velocity which is too far from the planet and escape velocity by using the energy conservation, we get

Rock’s initial velocity , v_{i}=2 v_{e}. Here the radius is R, so find the escape velocity as follows,

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            v_{e}^{2}=\frac{2 G M}{R}

            v_{e}=\sqrt{\frac{2 G M}{R}}

Where, M = Planet’s mass and G = constant.

From given conditions,

Surface potential energy can be expressed as,  U_{i}=-\frac{G M m}{R}

R tend to infinity when far away from the planet, so v_{f}=0

Then, kinetic energy at initial would be,

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Similarly, kinetic energy at final would be,

                k_{f}=\frac{1}{2} m v_{f}^{2}

Here, v_{f}=\text { final velocity }

Now, adding potential and kinetic energies of initial and final and equating as below, find the final velocity as

                 U_{i}+k_{i}=k_{f}+v_{f}

                 \frac{1}{2} m\left(2 v_{e}\right)^{2}-\frac{G M m}{R}=\frac{1}{2} m v_{f}^{2}+0

                  \frac{1}{2} m\left(2 v_{e}\right)^{2}-\frac{G M m}{R}=\frac{1}{2} m v_{f}^{2}

'm' and \frac{1}{2} as common on both sides, so gets cancelled, we get as

                   4\left(v_{e}\right)^{2}-\frac{2 G M}{R}=v_{f}^{2}

We know, v_{e}=\sqrt{\frac{2 G M}{R}}, it can be wriiten as \left(v_{e}\right)^{2}=\frac{2 G M}{R}, we get

                4\left(v_{e}\right)^{2}-\left(v_{e}\right)^{2}=v_{f}^{2}

                v_{f}^{2}=3\left(v_{e}\right)^{2}

Taking squares out, we get,

                v_{f}=\sqrt{3} v_{e}

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2 years ago
Jill puts her face in front of a convex mirror, 18 cm from the focal point of the mirror. If the focal point is located 12 cm fr
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A. 4 cm behind the mirror
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2 years ago
Case 1: A DJ starts up her phonograph player. The turntable accelerates uniformly from rest, and takes t₁ = 11.9 seconds to get
olga_2 [115]

Answer:

Part a)

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Part b)

N = 7.74 rev

Part c)

\alpha = 0.69 rad/s^2

Part d)

\alpha = 0.48 rad/s^2

Part e)

t = 9.14 s

Explanation:

Part a)

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\omega = 2\pi f

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\omega = 8.17 rad/s

Part b)

Since turn table is accelerating uniformly

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Part c)

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Part d)

When its angular speed changes to 120 rpm

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\omega_2 = 2\pi (\frac{120}{60})

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Part e)

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\omega_f - \omega_i = \alpha t

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