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gayaneshka [121]
3 years ago
10

Reclamation would be classified as a _____ human activity for the environment. Positive or negative?

Physics
1 answer:
Mice21 [21]3 years ago
7 0
This is could be both negative and positive but in this case I believe it’s referring to the reclamation of the environment in order to grow our agricultural
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You are designing a flywheel. It is to start from rest and then rotate with a constant angular acceleration of 0.200 rev/s^2. Th
Rama09 [41]

Answer:

 I = 8.75 kg m

Explanation:

This is a rotational movement exercise, let's start with kinetic energy

        K = ½ I w²

They tell us that K = 330 J, let's find the angular velocity with kinematics

      w² = w₀² + 2 α θ

as part of rest w₀ = 0

      w = √ 2α θ

let's reduce the revolutions to the SI system

      θ = 30.0 rev (2π rad / 1 rev) = 60π rad

let's calculate the angular velocity

      w = √(2  0.200  60π)

      w = 8.683 rad / s

we clear from the first equation

        I = 2K / w²

let's calculate

        I = 2 330 / 8,683²

        I = 8.75 kg m

4 0
3 years ago
Plz di all plz i will give brainest and thanks to best answer do it right
Inga [223]

Answer:

area 4

Explanation:

area 4 has low pressure

7 0
3 years ago
We dont see objects. We see the light ____ off objects.
Romashka [77]
The answer is BBBBBBBBB
4 0
3 years ago
This is an object's change in motion per unit time in a specified direction.
timurjin [86]

Answer:

Velocity

Explanation:

Velocity is an object's change in motion per unit time in a specified direction

7 0
3 years ago
Read 2 more answers
A metallic wire has a diameter of 4.12mm. When the current in the wire is 8.00A, the drift velocity is 5.40×10−5m/s.What is the
podryga [215]

Answer:

6.9\times 10^{28}m^{-3}

Explanation:

We are given that

Diameter of wire=d=4.12 mm

Radius of wire=rr=\frac{d}{2}=\frac{4.12}{2}=2.06mm=2.06\times 10^{-3} m

1mm=10^{-3} m

Current=I=8 A

Drift velocity=v_d=5.4\times 10^{-5} m/s

We have to find the density of free electrons in the metal

We know that

Density of electron=n=\frac{I}{v_deA}

Using the formula

Density of free electrons=\frac{8}{5.4\times 10^{-5}\times 1.6\times 10^{-19}\times 3.14\times (2.06\times 10^{-3})^2}

By using Area of wire=\pi r^2

\pi=3.14\\e=1.6\times 10^{-19} C

Density of free electrons=6.9\times 10^{28}m^{-3}

3 0
3 years ago
Read 2 more answers
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