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kiruha [24]
2 years ago
9

BRAINLIEST+10PTS: does ozone absorb just harmful UV rays? Or just a random portion of UV?

Physics
2 answers:
emmasim [6.3K]2 years ago
4 0

Answer:As sunlight passes through the atmosphere, all UVC and most UVB is absorbed by ozone, water vapour, oxygen and carbon dioxide. UVA is not filtered as significantly by the atmosphere. Is there a connection between ozone depletion and UV radiation? Ozone is a particularly effective absorber of UV radiation.

Explanation:

I had old notes

Umnica [9.8K]2 years ago
3 0

Answer:

The ozone layer absorbs a random portion of UV rays.

Explanation:

The ozone layer absorbs a random portion of UV rays because the only way UV rays could be harmful to your skin is if there is too much of it. All UV rays are bad, but the only reason you get harmed from them is if you absorb an excessive amount of them.

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What is the symbol for ammonium
rosijanka [135]
Symbol of ammonium is

3 0
3 years ago
What happens when you want to move the boat forward? You pull the oars toward yourself.Explain why you do this.
vampirchik [111]

Answer:

You pull on the oars. By the third law, the oars push back on your hands, but that’s irrelevant to the motion of the boat. The other end of each oar (the blade) pushes against the water. By the third law, the water pushes back on the oars, pushing the boat forward.

8 0
3 years ago
Read 2 more answers
8. An effort force of 15 Newtons is applied to an ideal pulley system to lift up a 16 Newton object. If the effort force is exer
Sonbull [250]

Answer:

the distance that the object is raised above its initial position is 5.625 m.​

Explanation:

Given;

applied effort, E = 15 N

load lifted by the ideal pulley system, L = 16 N

distance moved by the effort, d₁ = 6 m

let the distance moved by the object = d₂

For an ideal machine, the mechanical advantage is equal to the velocity ratio of the machine.

M.A = V.R

M.A = \frac{Load}{Effort} = \frac{L}{E} \\\\V.R = \frac{disatnce \ moved \  by \ the \ effort}{disatnce \ moved \  by \ the \ load} = \frac{d_1}{d_2} \\\\For \ ideal \ machine; \ M.A = V.R\\\\\frac{L}{E} = \frac{d_1}{d_2} \\\\d_2 = \frac{E \times d_1}{L} \\\\d_2 = \frac{15 \times 6}{16} \\\\d_2 = 5.625 \ m

Therefore, the distance that the object is raised above its initial position is 5.625 m.​

3 0
3 years ago
Anybody know these ?
pashok25 [27]

Answer:

1. a) 72 N.

2. a) 2 m/s².

Explanation:

Given the following data;

1. Mass = 90kg

Acceleration = 0.8 m/s²

To find the force;

Force = mass * acceleration

Force = 90 * 0.8

Force = 72 Newton.

2. Mass = 50kg

Force = 100N

To find the magnitude of acceleration;

Acceleration = force/mass

Acceleration = 100/50

Acceleration = 2 m/s²

5 0
3 years ago
Fiora starts riding her bike at 20 mi/h. After a while, she slows down to 14 mi/h, and maintains that speed for the rest of the
goblinko [34]

Answer:

3.5 hours

Explanation:

Speed = distance/time

Let the distance that Fiora biked at 20 mi/h through be x miles and the time it took her to bike through that distance be t hours at 20 mi/h

Then, the rest of the distance that she biked at 14 mi/h is (112 - x) miles

And the time she spent biking at 14 mi/h the rest of the distance = (6.5 - t) hours

Her first biking speed = 20 mph = 20 miles/hour

Speed = distance/time

20 = x/t

x = 20 t (eqn 1)

Her second biking speed = 14 mph = 14 miles/hour

14 = (112 - x)/(6.5 - t)

112 - x = 14 (6.5 - t)

112 - x = 91 - 14t (eqn 2)

Substitute for x in (eqn 2)

112 - 20t = 91 - 14t

20t - 14t = 112 - 91

6t = 21

t = 3.5 hours

x = 20t = 20 × 3.5 = 70 miles.

(112 - x) = 112 - 70 = 42 miles

(6.5 - t) = 6.5 - 3.5 = 3 hours

Meaning that she travelled at 20 mi/h for 3.5 hours.

4 0
3 years ago
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