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Maurinko [17]
2 years ago
13

A cement factory emits 900 kilograms of CO2 to produce 1,000 kilograms of cement. A fully grown tree removes six kilograms of CO

2 per day from the air. One acre of land can support 200 mature trees. A company builds a factory that will produce 100,000 kilograms of cement. To make the factory carbon neutral, the owners need to grow trees over an area of land measuring acres.
Physics
2 answers:
Dmitriy789 [7]2 years ago
6 0
<h3><u>Answer;</u></h3>

To make the factory carbon neutral, the owners need to grow<em><u> 15000 trees</u></em> over an area of land measuring <em><u>75 acres</u></em>

<h3><u>Explanation and solution;</u></h3>

From the information;

900 kg CO2 = 1000 kg Cement

1 tree = 6 kg CO2

1 acre = 200 trees

<em>100000 kg Cement will require;</em>

<em>=(900 × 100000)/1000</em>

<em>= 90,000 kg of CO2</em>

<em>But 1 tree = 6 kg of CO2</em>

<em>Number of trees = 90,000/6</em>

<em>                            = 15,000 trees </em>

<em>But, 1 acre = 200 trees</em>

<em>Number of acres = 15,000/200</em>

<em>                             = 75 acres of land </em>

Therefore;

To make the factory carbon neutral, the owners need to grow<em><u> 15000 trees</u></em> over an area of land measuring <em><u>75 acres</u></em>

Usimov [2.4K]2 years ago
5 0
Using the relations:

900 kg CO2 = 1000 kg Cement
1 tree = 6 kg CO2
1 acre = 200 trees

Given:

100000 kg Cement, how many acres of land should owners have to counteract the emission?

100000 kg Cement = 90000 kg CO2 = 15000 trees = 75 acres of land

Therefore, the owners should have 75 acres of land filled with trees to counteract the CO2 emission.
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Answer:

as its mass and velocity will less so its momentum will be less than that of baseball

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2 years ago
What is the value of g on the surface of Saturn? Assume M-Saturn = 5.68×10^26 kg and R-Saturn = 5.82×10^7 m.Choose the appropria
Likurg_2 [28]

Answer:

Approximately \rm 11.2 \; N \cdot kg^{-1} at that distance from the center of the planet.

Option A) The low value of g near the cloud top of Saturn is possible because of the low density of the planet.

Explanation:

The value of g on a planet measures the size of gravity on an object for each unit of its mass. The equation for gravity is:

\displaystyle \frac{G \cdot M \cdot m}{R^2},

where

  • G \approx 6.67\times 10^{-11}\; \rm N \cdot kg^{-2} \cdot m^2.
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  • m is the mass of the object.

To find an equation for g, divide the equation for gravity by the mass of the object:

\displaystyle g = \left.\frac{G \cdot M \cdot m}{R^2} \right/\frac{1}{m} = \frac{G \cdot M}{R^2}.

In this case,

  • M = 5.68\times 10^{26}\; \rm kg, and
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Calculate g based on these values:

\begin{aligned} g &= \frac{G \cdot M}{R^2}\cr &= \frac{6.67\times 10^{-11}\; \rm N \cdot kg^{-2} \cdot m^2\times 5.68\times 10^{26}\; \rm kg}{\left(5.82\times 10^7\; \rm m\right)^2} \cr &\approx 11.2\; \rm N\cdot kg^{-1} \end{aligned}.

Saturn is a gas giant. Most of its volume was filled with gas. In comparison, the earth is a rocky planet. Most of its volume was filled with solid and molten rocks. As a result, the average density of the earth would be greater than the average density of Saturn.

Refer to the equation for g:

\displaystyle g = \frac{G \cdot M}{R^2}.

The mass of the planet is in the numerator. If two planets are of the same size, g would be greater at the surface of the more massive planet.

On the other hand, if the mass of the planet is large while its density is small, its radius also needs to be very large. Since R is in the denominator of g, increasing the value of R while keeping M constant would reduce the value of g. That explains why the value of g near the "surface" (cloud tops) of Saturn is about the same as that on the surface of the earth (approximately 9.81\; \rm N \cdot kg^{-1}.

As a side note, 5.82\times 10^7\rm \; m likely refers to the distance from the center of Saturn to its cloud tops. Hence, it would be more appropriate to say that the value of g near the cloud tops of Saturn is approximately \rm 11.2 \; N \cdot kg^{-1}.

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Explanation:

The given data  is as follows.

Mass of oxygen present = 100 mg = 100 \times 10^{-3} g

So, moles of oxygen present are calculated as follows.

      n = \frac{100 \times 10^{-3}}{32}

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Diameter of cylinder = 6 cm = 6 \times 10^{-2} m

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Now, we will calculate the cross sectional area (A) as follows.

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Length of tube = 11 cm = 0.11 m

Hence, volume (V) = 2.82 \times 10^{-3} \times 0.11

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Now, we assume that the inside pressure is P .

And,   P_{atm} = 100 kPa = 100000 Pa,

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Hence, force required to open is as follows.

      Force = Pressure difference × A

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We are given that force is 173 N.

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At time t, gives the position of a 3.0 kg particle relative to the origin of an xy coordinate system ( ModifyingAbove r With rig
Elena-2011 [213]

Complete Question

  The complete Question is shown on the first uploaded image

Answer:

a

The torque acting on the particle is  \tau = 48t \r k

b

The magnitude of the angular momentum increases relative to the origin

Explanation:

From the equation we are told that

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       The mass of the particle is m = 3.0 kg

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where \r l is change in angular momentum which is mathematically represented as

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Where X mean cross- product

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Substituting for  \r r

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The magnitude of the angular momentum can be evaluated mathematically as

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Answer:

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