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Neko [114]
3 years ago
5

What is the Ozone layer?

Physics
2 answers:
yaroslaw [1]3 years ago
4 0
"<span>a layer in the earth's stratosphere at an altitude of about 6.2 miles (10 km) containing a high concentration of ozone, which absorbs most of the ultraviolet radiation reaching the earth from the sun."

Hope this helps!
</span>
djverab [1.8K]3 years ago
4 0
"The ozone layer is the layer in the upper atmosphere that contains a higher concentration of ozone than the rest of the atmosphere"
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A 78-kg skydiver has a speed of 62 m/s , what is the kinetic energy?​
Cloud [144]

Answer:

149,916J

Explanation:

Pneumonic device: Kevin is half-mad and very square

this translates to: KE=(1/2)mv^2 !!

KE=(1/2)(78)(62^2)

KE=(39)(3844)

KE=149,916 Joules

3 0
3 years ago
Read 2 more answers
Position vs Time
Annette [7]

Answer:

Determining the Slope on a p-t Graph. It was learned earlier in Lesson 3 that the slope of the line on a position versus time graph is equal to the velocity of the object. If the object is moving with a velocity of +4 m/s, then the slope of the line will be +4 m/s

7 0
3 years ago
Suppose you dissolve 4.75 g of another solid (not urea) in 50.0 mL of water. You note that the temperature changes from 25.0 °C
torisob [31]

Answer:

Mass of the solution  = 54.75 g

Explanation:

Mass of solid dissolved = 4.75 g

Mass of the solution = Mass of solid dissolved + Mass of water.

Mass of water = Volume x Density

Volume = 50 mL = 50 cm³

Density = 1 g/cm³

Mass of water = 50 x 1 = 50 g

Mass of the solution = 4.75 + 50 = 54.75 g

8 0
4 years ago
7. What is the change in gravitational potential energy for a 120.80-kg giant walking down a hill from a
ratelena [41]

Answer:

Explanation:

It takes 1200 J of work to lift a car high enough to change a tire.

4 0
3 years ago
Use the ratio version of Kepler’s third law and the orbital information of Mars to determine Earth’s distance from the Sun. Mars
frez [133]

Answer:

<em>1.49 x </em>10^{11}<em></em>

<em></em>

Explanation:

Kepler's third law states that <em>The square of the orbital period of a planet is directly proportional to the cube of its orbit.</em>

Mathematically, this can be stated as

T^{2} ∝ R^{3}

<em>to remove the proportionality sign we introduce a constant</em>

T^{2} = kR^{3}

k = \frac{T^{2} }{R^{3} }

Where T is the orbital period,

and R is the orbit around the sun.

For mars,

T = 687 days

R = 2.279 x 10^{11}

for mars, constant k will be

k = \frac{687^{2} }{(2.279*10^{11}) ^{3} } = 3.987 x 10^{-29}

For Earth, orbital period T is 365 days, therefore

365^{2} = 3.987 x 10^{-29} x R^{3}

R^{3} = 3.34 x 10^{33}

R =<em> 1.49 x </em>10^{11}<em></em>

8 0
3 years ago
Read 2 more answers
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