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lesya692 [45]
3 years ago
6

The atmosphere's layers

Physics
2 answers:
Rasek [7]3 years ago
6 0

Answer:

1. The lowest stratum of the atmosphere

troposphere

2. Most of the earth's ozone is located in the  

stratosphere

3. Earth coldest temperature occur in the  

mesosphere

Explanation:

Genrish500 [490]3 years ago
6 0

1: Troposphere

The troposphere is the lowest layer of the Earth's atmosphere. The air is very well mixed and the temperature decreases with altitude. Air in the troposphere is heated from the ground up. The surface of the Earth absorbs energy and heats up faster than the air does.

2: Stratosphere

Ozone is mainly found in two regions of the Earth's atmosphere. Most ozone (about 90%) resides in a layer that begins between 6 and 10 miles (10 and 17 kilometers) above the Earth's surface and extends up to about 30 miles (50 kilometers). This region of the atmosphere is called the stratosphere.

3: Stratosphere

As the mesosphere extends upward above the stratosphere, temperatures decrease. The coldest parts of our atmosphere are located in this layer and can reach –90°C. In the forth layer from Earth's surface, the thermosphere, the air is thin, meaning that there are far fewer air molecules.

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The acceleration of the body is 2 m/s^2 while the deceleration is - 1.2 m/s^2.

<h3>What is the acceleration?</h3>

Let us recall that the acceleration is the change in the speed of a body with time. We have been told that the body accelerates for 3s and then decelerates to 2s. This implies that the total time that the object spent in motion is 5 s.

Thus;

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7 0
2 years ago
?a wire is stretched 30% what is the percentage change in resistance ​
Marat540 [252]

Answer:

The percentage change in resistance of the wire is 69%.

Explanation:

Resistance of a wire can be determined by,

R = (ρl) ÷ A

Where R is its resistance, l is the length of the wire, A its cross sectional area and ρ its resistivity.

When the wire is stretched, its length and area changes but its volume and resistivity remains constant.

l_{o} = 1.3l, and A_{o} = \frac{A}{1.3}

So that;

R_{o} = (ρl_{o}) ÷ A_{o} = (ρ × 1.3l) ÷ (\frac{A}{1.3})

    = (1.3lρ) ÷ (\frac{A}{1.3})

    = (1.3)^{2} × [(ρl) ÷ A]

   = 1.69R               (∵ R = (ρl) ÷ A)

R_{o} = 1.69R

Where R_{o} is the new resistance, l_{o} is the new length, and A_{o} is the new area after stretching the wire.

The change in resistance of the wire = R_{o} - R

                                      = 1.69R  - 1R

                                      = 0.69R

The percentage change in resistance = \frac{0.69R}{R} × 100

                                                               = 0.69 × 100

                                                              = 69%

The percentage change in resistance of the wire is 69%.

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