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MAXImum [283]
2 years ago
12

Consider the possible scenarios and determine which of them are true for positive feedbacks or negative feedbacks.A. Increased p

lant photosynthesis resulting from a temperature rise can remove carbon dioxide from the atmosphere which can potentially lead to positive feedback. B. Increased evaporation resulting from a temperature rise can produce more clouds which can potentially lead to a negative feedback. C. Increased plant photosynthesis resulting from a temperature rise can remove carbon dioxide from the atmosphere which can potentially lead to a negative feedback. D. Increased evaporation resulting from a temperature rise can produce more clouds which can potentially lead to a positive feedback. E. If ice cover melts due to a temperature rise the result will be decreased albedo or reflectivity which can potentially lead to a negative feedback. F. If ice cover melts due to a temperature rise the result will be decreased albedo or reflectivity which can potentially lead to a positive
Physics
1 answer:
avanturin [10]2 years ago
5 0

Answer:

Scenario A, B and E is True.

Explanation:

Scenario A) True. Removing carbon dioxide from atmosphere decreases greenhouse effect of atmosphere. Thus, temperature rise decreases.

Scenario B) True. The more evaporation creates the more greenhouse effect. Therefore, temperature rise increases.

Scenario C) False. Removing carbon dioxide from atmosphere decreases greenhouse effect of atmosphere. Thus, temperature rise decreases.

Scenario D) False. The more evaporation creates the more greenhouse effect. Therefore, temperature rise increases.

Scenario E) True. If reflected radiation increases from Earth, temperature rise of the Earth will decrease. Ice cover increases reflectivity which leads temperature level decrease.

Scenario F) False. If reflected radiation increases from Earth, temperature rise of the Earth will decrease. Ice cover increases reflectivity which leads temperature level decrease.

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If a current of 2.4 a is flowing in a cylindrical wire of diameter 2.0 mm, what is the average current density in this wire?
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The average current density in the wire is given by:

J=\frac{I}{A}

where I is the current intensity and A is the cross-sectional area of the wire.


The cross-sectional area of the wire is given by:

A=\pi r^2

where r is the radius of the wire. In this problem, r=\frac{d}{2}=\frac{2.0 mm}{2}=1.0 mm=0.001 m, so the cross-sectional area is

A=\pi (0.001 m)^2=3.14 \cdot 10^{-6} m^2


and the average current density is

J=\frac{I}{A}=\frac{2.4 A}{3.14 \cdot 10^{-6} m^2}=7.64 \cdot 10^5 A/m^2

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2 years ago
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A 10-kilogram box is at static equilibrium, and the downward pull of gravity acting on the box is 98 newtons. What is the minimu
Nikolay [14]
The box is at equilibrium, so the net force on the box is zero (the force of gravity on the box is equal to the force exerted up on the box by the surface on which it rests.)
To pick up the box, our upward force must be greater than the force of gravity on the box (the weight). So, we must lift up the box with a force greater than 98 newtons. :)
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If the force of gravity between a book of mass 0.50 kg and a calculator of 0.100 kg is 1.5 × 10-10 N, how far apart are they?  (
valkas [14]
The gravitational force between two masses m₁ and m₂ is
F=G \frac{m_{1} m_{2}}{d^{2}}
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G = 6.67408 x 10⁻¹¹ m³/(kg-s²), the gravitational constant
d =  distance between the masses.

Given:
F = 1.5 x 10⁻¹⁰ N
m₁ = 0.50 kg
m₂ = 0.1 kg

Therefore
1.5 x 10⁻¹⁰ N = (6.67408 x 10⁻¹¹ m³/(kg-s²))*[(0.5*0.1)/(d m)²]
d² = [(6.67408x10⁻¹¹)*(0.5*0.1)]/1.5x10⁻¹⁰
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d = 0.1492 m = 149.2 mm

Answer: 149.2 mm
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2 years ago
A pole-vaulter is nearly motionless as he clears the bar, set 4.2 m above the ground. he then falls onto a thick pad. the top of
scZoUnD [109]
Refer to the diagram shown below.

Neglect wind resistance, and use g = 9.8 m/s².

The pole vaulter falls with an initial vertical velocity of u = 0.
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The pole vaulter comes to res after the pad compresses by  50 cm (or 0.5 m).
If the average acceleration (actually deceleration) is (a m/s²), then
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Answer: - 82 m/s² (or a deceleration of 82 m/s²)

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