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Lubov Fominskaja [6]

3 years ago

6

Why one part of the earth's surface is arid and dry and a nearby part is lush and wet is explained by the term?

1 answer:

Darina [25.2K]3 years ago

8 0

The process that explains why one part of the earth's surface is arid and dry and a nearby part is lush and wet is areal differentiation. It is<span> an approach to geography that shows </span>the dependence of the distribution of physical and human phenomena and the relation to each other from the physical location. Areal integration on the other hand is the approach that studies how places interact with each other.

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If the magnetic field of an electromagnetic wave is in the +x-direction and the electric field of the wave is in the +y directio

alexandr1967 [171]

Answer:

Explanation:

The direction of propagation of electromagnetic wave

is given by the direction of vector E x B where E is electrical field , B is magnetic field .

Given Electric field = E i because it is along x axis

Magnetic field = Bj because it is along y axis

E x B = Ei x Bj

= EB k .

so direction of E x B is along k direction or z - axis so wave is propagating along z - axis .

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3 years ago

When Arti kicks a football, two forces interact. Arti's foot exerts a force on the ball. What exerts a force on Arti's foot?

vagabundo [1.1K]

The football and air resistance between contact

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3 years ago

When do sunspots disappear?

dedylja [7]

Answer:

In 5 years or so, the sun will be awash in sunspots and more prone to violent bursts of magnetic activity.

Explanation

once the magnetic field weakens the area and cold plasma enters the area of the sunspot

8 0

2 years ago

At a certain location, a gravitational force with a magnitude of 350 newtons acts on a 70.-kilogram astronaut. What is the magni

creativ13 [48]

Answer:

3. 5.0N/kg

Explanation:

Gravitational field strength = gravitational force/mass of astronaut = 350N/70kg = 5.0N/kg

6 0

3 years ago

A thin ring of radius 73 cm carries a positive charge of 610 nC uniformly distributed over it. A point charge q is placed at the

kow [346]

Answer:

q = - 93.334 nC

Explanation:

GIVEN DATA:

Radius of ring 73 cm

charge on ring 610 nC

ELECTRIC FIELD p FROM CENTRE IS AT 70 CM

E = 2000 N/C

Electric field due tor ring is guiven as

$E = \frac{KQx}{[x^2+ R^2]^{3/2}}$

$E = \frac{9\time 10^9 \times 610\times 10^[-9} 0.70}{(0.70^2 + 0.73^2)^{3/2}}$

E1 = 3714.672 N/C

electric field due to point charge q

$E =\frac[kq}{x^2}$

$E = \frac{9\times 10^9 \times q}{0.70^2}$

$E2 = 1.837\times 10^{10}\times q$

now the eelctric charge at point P is

E = E1 + E2 $2000 = 3714.672 + 1.837\times 10[10} \times q$

solving for q

q = - 93.334 nC

7 0

3 years ago