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harkovskaia [24]
2 years ago
13

What are the layers of the Earth's atmosphere?

Physics
2 answers:
Mandarinka [93]2 years ago
4 0
<span>Layers of the Earth's Atmosphere: Troposphere, Stratosphere, Mesosphere, Thermosphere, and Exosphere - Windows to the Universe.</span>
borishaifa [10]2 years ago
3 0
<span> <span>Layers of the Earth's Atmosphere: Troposphere, Stratosphere, Mesosphere, Thermosphere, and Exosphere - Windows to the Universe.
</span> </span> Comments
<span>0</span>
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Why is the work output of a machine never equal to the work input? A. Some work input is used to overcome friction. B. Input dis
SVETLANKA909090 [29]
The answer is A. <span>Some work input is used to overcome friction. </span>
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3 years ago
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Nancy immediately remembers she had promised to return her grandmother's call from earlier this morning. What type of memory is
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I believe its Procedural Memory
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3 years ago
Several wires of varying thickness are all made of the same material and all have the same length. If the wires are arranged in
aev [14]

Answer:

Option A is correct.

The wires will be arranged in order of increasing resistance.

Explanation:

The resistance of a wire is given by

r = (ρl)/A

where r = resistance of the wire

ρ = resistivity of the wire

L = length of the wire

A = cross sectional area of the wire

Provided all the other parameters are constant, resistance is inversely proportional to cross sectional area

r ∝ (1/A)

And the the cross sectional Area of the wire increases with increase in thickness & decreases with thickness

So, decreasing thickness ----> Decreasing Cross sectional Area ----> Increasing resistance.

5 0
3 years ago
Paul’s 10 kg baby sister Susan sits on a mat. Paul pulls the mat across the floor using a rope that is angled 30° above the floo
kiruha [24]

Answer:

The speed of Susan is 2.37 m/s

Explanation:

To visualize better this problem, we need to draw a free body diagram.

the work is defined as:

W=F*d*cos(\theta)

here we have the work done by Paul and the friction force, so:

W_p=F_p*d*cos(0)\\F_p=30N*cos(30^o)=26N\\W_p=26*3*(1)=78J

W_f=F_f*d*cos(180)\\F_f=\µ*(10*9.8-30N*sin(30^o))=16.6N\\W_p=16.6*3*(-1)=50J

Now the change of energy is:

W_p-W_f=\frac{1}{2}m*v^2\\v=\sqrt{\frac{2(78J-50J)}{10kg}}\\v=2.37m/s

4 0
3 years ago
Read 2 more answers
How can acceleration be changed without changing speed?
katrin [286]
Since velocity is a speed and a direction, there are only two ways for you to accelerate: change your speed or change your direction—or change both. If you're not changing your speed and you're not changing your direction, then you simply cannot be accelerating—no matter how fast you're going.
3 0
2 years ago
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