3 years ago

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

1 answer:

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

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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

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3 years ago

We dont see objects. We see the light ____ off objects.

Romashka [77]

The answer is BBBBBBBBB

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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

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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=r $r=\frac{d}{2}=\frac{4.12}{2}=2.06mm=2.06\times 10^{-3} m$

$1mm=10^{-3} m$

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

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