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krok68 [10]
2 years ago
12

Cutting down trees can decrease the amount of water vapor entering the air through the processes of ____?

Physics
1 answer:
AlexFokin [52]2 years ago
5 0

✿  Transpiration is the process by which water is carried through plants from roots and release water vapor through their leaves.

✿  Cutting down trees can decrease the amount of water vapor entering the air through the processes of <u>Transpiration</u>

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_____ is the kinetic energy of an object, proportional to the object's moment of inertia and the square of its angular velocity.
Juli2301 [7.4K]
Rotational kinetic energy <span>is the kinetic energy of an object, proportional to the object's moment of inertia and the square of its angular velocity.</span>
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3 years ago
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Irina-Kira [14]

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7 0
2 years ago
A solid conducting sphere (radius = 5.0 cm) has a charge of 0.25 nC distributed uniformly on its surface. If point A is located
nadezda [96]

Answer:

\Delta V = 30V

Explanation:

give data:

inside diameter = 5.0 cm

charge q = 0.25 nC

Outside diameter = 15 cm

potential V at inside sphere is = V=\frac{kq}{R}

potential V at outside sphere is = V=\frac{kq}{r}

k is constant whose value is = 8.99 *10^{9}Nm^{2}/c^2

then potential difference between two point is

\Delta V = kq \left [\frac{1}{R}-\frac{1}{r}  \right ]

\Delta V = 9*10^{9}*0.25*10^{-9} \left [\frac{1}{0.05}-\frac{1}{0.15} \right ]

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6 0
3 years ago
How the choice of building materials and techniques can help them cool down faster, after heating up.
Sauron [17]

Answer:

The choice of building materials and techniques can help them cool down faster, after heating up is explained below in brief details.

Explanation:

Building materials

Buildings that are composed of rock, bricks, or pavement, or inserted into the territory, can seem cooler thanks to the great "thermal mass" of these elements – that is, their capability to assimilate and discharge heat gently, thereby softening temperatures over time, producing daytime cooler and night-time warmer.

3 0
3 years ago
Two crates, 3kg and 5 kg respectively, are connected with a thin rope.they are pulled up a rough plane which is inclined at 30de
Makovka662 [10]

Answer:

F = 85.24 N

T = 30.84 N

Explanation:

The parameters given are

Mass of crates = m₁ = 3 kg and m₂ = 5 kg

Component of masses acting along the plane = mgsin(θ)

Which gives;

F - (3×9.81×sin(30) + 10 + 5×9.81×sin(30) + 17) = m×a

So that we have;

2 = (F - 66.24)(3 + 5)

F = 16 + 66.24 = 85.24 N

The tension T between the crates = F × m₁/(m₁+m₂) = 85.24 × 3/(3+5) = 30.84 N

6 0
3 years ago
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