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koban [17]
3 years ago
7

MATCH THE FOLLOWING

Physics
2 answers:
olchik [2.2K]3 years ago
8 0

shdhdggxshjwhwywsd dccefc

lesantik [10]3 years ago
3 0

Answer:

afrcrftftgyfygygygygygygyvyvuguvy5vyvycycy

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Why are most of the world’s deserts located between latitudes 10°n to 30°n and 10°s to 30°s.
RoseWind [281]

The bulk of the world's deserts are located at 30 degrees north latitude and 30 degrees south latitude, when the warm equatorial air begins to descend. The heavy, warm, descending air vaporises large amounts of water from the ground's surface. As a result, the environment is rather dry.

<h3>Why are the majority of the desert regions on Earth located between 20 and 30 degrees latitude?</h3>

The zones of falling air are those between 20 and 30 latitudes on the western borders of continents (high pressure and dry weather). As a result, the moisture continues to decrease as the air is compressed and warmed as it falls.

Where the scorching equatorial air starts to descend, the majority of the world's deserts are found between 30 degrees north and 30 degrees south latitude. Large volumes of water are vaporised off the surface of the ground by the thick, warming, falling air. As a result, the climate is extremely dry.

Learn more about latitude refer

brainly.com/question/1939015

#SPJ4

3 0
1 year ago
I WILL GIVE BRAINLIEST....<br>Determine the value of F...​
aleksley [76]
F should be 10. If The Whole top is 50cm, then we should subtract 10n and 30n which gives us 10.

Or it could be 15 if both top and bottom are 25. 10+15= 25.
6 0
3 years ago
What term describes the resistance that one encounters when moving over another
Anvisha [2.4K]
Friction? For example, like when a car's tires skid on rough concrete.
5 0
3 years ago
A 5-cm-high peg is placed in front of a concave mirror with a radius of curvature of 20 cm.
Andrei [34K]

Answer:

Explanation:

Using the magnification formula.

Magnification = Image distance(v)/object distance(u) = Image Height(H1)/Object Height(H2)

M = v/u = H1/H2

v/u = H1/H2...1

3) Given the radius of curvature of the concave lens R = 20cm

Focal length F = R/2

f = 20/2

f = 10cm

Object distance u = 5cm

Object height H2= 5cm

To get the image distance v, we will use the mirror formula

1/f = 1/u+1/v

1/v = 1/10-1/5

1/v = (1-2)/10

1/v =-1/10

v = -10cm

Using the magnification formula

(10)/5 = H1/5

10 = H1

H1 = 10cm

Image height of the peg is 10cm

4) If u = 15cm

1/v = 1/f-1/u

1/v = 1/10-1/15

1/v = 3-2/30

1/v = 1/30

v = 30cm

30/15 = H1/5

15H1 = 150

H1/= 10cm

5) if u = 20cm

1/v = 1/f-1/u

1/v = 1/10-1/20

1/v = 2-1/20

1/v = 1/20

v = 20cm

20/20 = H1/5

20H1 = 100

H1 = 5cm

6) If u = 30cm

1/v = 1/f-1/u

1/v = 1/10-1/30

1/v = 3-1/30

1/v = 2/30

v = 30/2 cm

v =>15cm

15/30 = Hi/5

30H1 = 75

H1 = 75/30

H1 = 2.5cm

4 0
3 years ago
How long does GSR last ?
SSSSS [86.1K]

Answer:

i think that the answer might be be C. one day. because it last about 4 to 6 hours

4 0
3 years ago
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