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3 years ago

Positive feedback interactions in earth’s systems are always a result of human action. T/F

Physics

1 answer:

VLD [36.1K]3 years ago

4 0

Positive feedback interactions in earth’s systems are always a result of human action is a FALSE statement.

Explanation:

Earth is a unstable equilibrium which tends to move out of equilibrium, but several other factors try to bring it back in equilibrium again. Earth has a different actions going on both on its surface and also inside it.

Human can alter, or can modify a very small part of the events that occur on earth’s surface. But they don’t have any control on what’s going inside earth’s core. So positive feedback interactions are not only the results of human interaction, but also different other factors.

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Learning Goal: To practice Problem-Solving Strategy 7.2 Problems Using Mechanical Energy II. The Great Sandini is a 60.0-kg circ

andreev551 [17]

Answer:

v = 15.8 m/s

Explanation:

Let's analyze the situation a little, we have a compressed spring so it has an elastic energy that will become part kinetic energy and a potential part for the man to get out of the barrel, in addition there is a friction force that they perform work against the movement. So the variation of mechanical energy is equal to the work of the fictional force

$W_{fr}$ = ΔEm = $Em_{f}$ -Em₀

Let's write the mechanical energy at each point

Initial

Em₀ = Ke = ½ k x²

Final

$Em_{f}$ = K + U = ½ m v² + mg y

Let's use Hooke's law to find compression

F = - k x

x = -F / k

x = 4400/1100

x = - 4 m

Let's write the energy equation

fr d = ½ m v² + mgy - ½ k x²

Let's clear the speed

v² = (fr d + ½ kx² - mg y) 2 / m

v² = (40 4.00 + ½ 1100 4² - 60.0 9.8 2.50) 2/60.0

v² = (160 + 8800 - 1470) / 30

v = √ (229.66)

v = 15.8 m/s

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3 years ago

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The planet should move around the elliptical orbit, and two segments of the orbit should become shaded in green. What aspect(s)

strojnjashka [21]

Answer: not sure

Explanation:

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2 years ago

Elaborate on the classification of aluminum. A) Aluminum is a noble gas and has a full valence electron shell, is chemically non

scZoUnD [109]

The correct option is (B) Aluminum is a metal and is shiny, malleable, ductile, conducts heat and electricity, forms basic oxides, and forms cations in aqueous solution.

Since Aluminium is in group 13, and all the elements in group 13 are either metals or metalloids(Boron). Hence we are left with option (B) and (D). Boron is the only metalloid in group 13 and aluminium is a metal(not a metalloid); therefore, we are left with only one option which is Option (B). And Aluminium is shiny, malleable, ductile, conducts heat and electricity, forms basic oxides, and forms cations in aqueous solution.

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3 years ago

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A point charge q1 = 1.0 µC is at the origin and a point charge q2 = 6.0 µC is on the x axis at x = 1 m.

iris [78.8K]

To solve this problem we will apply the concepts related to the Electrostatic Force given by Coulomb's law. This force can be mathematically described as

$F = \frac{kq_1q_2}{d^2}$

Here

k = Coulomb's Constant

$q_{1,2} =$ Charge of each object

d = Distance

Our values are given as,

$q_1 = 1 \mu C$

$q_2 = 6 \mu C$

d = 1 m

$k = 9*10^9 Nm^2/C^2$

a) The electric force on charge $q_2$ is

$F_{12} = \frac{ (9*10^9 Nm^2/C^2)(1*10^{-6} C)(6*10^{-6} C)}{(1 m)^2}$

$F_{12} = 54 mN$

Force is positive i.e. repulsive

b) As the force exerted on $q_2$ will be equal to that act on $q_1$ ,

$F_{21} = F_{12}$

$F_{21} = 54 mN$

Force is positive i.e. repulsive

c) If $q_2 = -6 \mu C$ , a negative sign will be introduced into the expression above i.e.

$F_{12} = \frac{(9*10^9 Nm^2/C^2)(1*10^{-6} C)(-6*10^{-6} C)}{(1 m)^{2}}$

$F_{12} = F_{21} = -54 mN$

Force is negative i.e. attractive

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2 years ago

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N76 [4]

Answer:

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Explanation:

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3 years ago