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2 years ago

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

2 answers:

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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3 years ago

Which unit abbreviation is a measurement of force?

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3 years ago

A person pulls a crate of mass M = 63 kg a distance 40.0 m along a horizontal floor by a constant force FP = 130 N, which acts a

GrogVix [38]

Answer:

Check attachment for solution and diagrams

Explanation:

Given that,

Mass of crate m=63kg

Distance travelled d=40m

Horizontal force Fx=130N

Angle the force applied on cord makes with horizontal is θ=23°.

The weight of the crate is given by

W=mg

W=63×9.81

W=618.03N

Horizontal force Fx=130N

Resolving the applied Force F to the horizontal will give

Fx=FCos θ

F=Fx/Cos θ

F=130/Cos23

F=141.2N

a. Check attachment for model diagram

b. Check attachment for free body diagram

c. Check attachment for pictorial representation

d. Work done by gravitational force.

We, know that the body did not move upward, then the distance d=0

Work done is given as

W=F×d

So, d=0

W=F×0

W=0J

So, no work is done by gravity

e. Normal force?

Using newton law of motion

ΣFy = may

Since the body is not moving upward, then ay=0m/s²

N+141.2Sin23-618.03=0

N=618.03-141.2Sin23

N=562.86N

f. Work done by normal force.

The body is not moving upward, then the distance is zero

d=0

Work done by normal=normal force × distance

Wn=562.86×0

Wn=0J

No work is done by the normal force

g. Frictional force?

Since the coefficient of kinetic friction is zero, then the surface is frictionless

So, no frictional force is acting on the body

Fictional force is given as

Fr=μk•N

Given that, μk=0

Fr=0×562.86

Fr=0N

d. Work done by frictional force?

Since the frictional force Is zero, then, no work is done by friction

W(friction ) = frictional force × d

Here, the body moved a distance of 40m

W(fr)=0×40

W(fr)=0J

No work is done by friction

I. Work done by exerted force

The horizontal component of the exerted force is 130N and the body traveled a distance of 40m

Then, work done is given as

Workdone=force ×distance

Work done=130×40

W=5200J

W=5.2KJ

h. Net workdone?

Since no work is lost by friction, then, the net work done is equal to the work done by the exerted force.

Went, = work done by force exerted - work done by friction

Wnet=5200-0

Wnet, =5200

Wnet=5.2KJ

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3 years ago