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lys-0071 [83]
4 years ago
9

The sinking of dry air results in the formation of what terrestrial biome?

Physics
1 answer:
blondinia [14]4 years ago
8 0
The sinking of dry air results in the formation of deserts. Dry air would absorb water from the surrounding area. Since the air moves, it would carry this moisture away. Therefore the dry air produces little precipitation and removes any available moisture from the area producing a desert. This typically happens at about 30 degrees north and 30 degrees south latitude. Hot moist air rises at the equator. The water condenses and falls as rain. As the air moves north and south it becomes dry and sinks at approximately 30 degrees north and south. This dry air allows deserts to be formed.
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Which one of the following accurately describes the force of gravity?
elena55 [62]
Choice ' C ' is a true statement.
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3 years ago
A −3.0 nC charge is on the x-axis at x=−9 cm and a +4.0 nC charge is on the x-axis at x=16 cm. At what point or points on the y-
alexdok [17]

Answer:

y = 10.2 m

Explanation:

It is given that,

Charge, q_1=-3\ nC

It is placed at a distance of 9 cm at x axis

Charge, q_2=+4\ nC

It is placed at a distance of 16 cm at x axis

We need to find the point on the y-axis where the electric potential zero. The net potential on y-axis is equal to 0. So,

\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}=0

Here,

r_1=\sqrt{y^2+9^2} \\\\r_2=\sqrt{y^2+15^2}

So,

\dfrac{kq_1}{r_1}=-\dfrac{kq_2}{r_2}\\\\\dfrac{q_1}{r_1}=-\dfrac{q_2}{r_2}\\\\\dfrac{-3\ nC}{\sqrt{y^2+81} }=-\dfrac{4\ nC}{\sqrt{y^2+225} }\\\\3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}

Squaring both sides,

3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}\\\\9(y^2+225)=16\times (y^2+81)\\\\9y^2+2025=16y^2-+1296\\\\2025-1296=7y^2\\\\7y^2=729\\\\y=10.2\ m

So, at a distance of 10.2 m on the y axis the electric potential equals 0.

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Mental health focuses on which component?
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Which kind of energy does the soccer player transfer to the ball?
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3 years ago
The slider of mass m is released from rest in position A and slides without friction along the vertical-plane guide shown. Deter
Anuta_ua [19.1K]

The value of normal force as the slider passes point B is

  • 6 mg

The value of h when the normal force is zero

  • 3R/2

<h3>How to solve for the normal force</h3>

The normal force is calculated using the work energy principle which is applied as below

K₁ + U₁ = K₂

k represents kinetic energy

U represents potential energy

the subscripts 1,2 , and 3 = a, b, and c

for 1 to 2

K₁ + W₁ = K₂

0 + mg(h + R) = 0.5mv²₂

g(h + R) = 0.5v²₂

v²₂ = 2g(1.5R + R)

v²₂ = 2g(2.5R)

v²₂ = 5gR

Using summation of forces at B

Normal force, N  = ma + mg

N = m(a + g)

N = m(v²₂/R + g)

N = m(5gR/R + g)

N = 6mg

for 1 to 3

K₁ + W₁ = K₃ + W₃

0 + mgh = 0.5mv²₃ + mgR

gh = 0.5v²₃ + gR

0.5v²₃ = gh - gR

v²₃ = 2g(h - R)

at C

for normal force to be zero

ma = mg

v²₃/R = g

v²₃ = gR

and v²₃ = 2g(h - R)

gR = 2gh - 2gR

gR + 2gR = 2gh

3gR = 2gh

3R/2 = h

Learn more about normal force at:

brainly.com/question/20432136

#SPJ1

8 0
1 year ago
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