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3 years ago

How does upwelling affect the weather of a coastal region?

Physics

2 answers:

Bezzdna [24]3 years ago

8 0

Answer:A

Explanation:

Because if you look a the sentences it tells u but you have to look into more

SIZIF [17.4K]3 years ago

4 0

c.The warm surface water results in moist air and more rainfall.

Explanation:

During upwelling, cold water in the ocean is stirred up and brought to the surface.
The warmer surface water is then taken into deeper parts of the ocean.
Upwelling allows for nutrient mixing in the ocean and allows for useful gases to circulate well.
The warm surface water causes the air to be moisty.
When the air is carried landward towards the coast, it leads to rainfall when the saturated air releases the water.
The air then becomes cold and dry and it rises up.
Therefore, warm surface water results in moist air and more rainfall.

Learn more:

Ocean current brainly.com/question/4117397

#learnwithBrainly

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What is the angular speed of (a) the second hand, (b) the minute hand, and (c) the hour hand of a smoothly running analog watch?

Agata [3.3K]

A)We know the formula of the angular speed is ω = 2π / TWhere T is the time period.When second hand completes one revolution then the time taken is 60s.So T = 60sThen the angular speed of the second hand is ω= 2π / (60s) = 0.1047 rad/sb)When the minute hand completes one revolution the time taken is T = 1 hr = 3600sThen the angular speed of the minute hand is ω =(2π) / (3600s) = 0.001745 rad/sc)When the hour hand completes one revolution then the timeperiod is T = 12hrs = (12)(3600)sThen the angular speed of the hour hand is ω =(2π) / [(12)(3600)s] = 1.45444 x 10^-4 rad/s

8 0

3 years ago

Medical cyclotrons need efficient sources of protons to inject into their center. In one kind of ion source, hydrogen atoms (i.e

antoniya [11.8K]

Answer:

$r=5.278\times 10^{-4}\ m$

Explanation:

Given that:

magnetic field intensity, $B=0.07\ T$

kinetic energy of electron, $KE=1.2\ eV= 1.2\times 1.6\times 10^{-19}\ J= 1.92\times 10^{-19}\ J$

we have mass of electron, $m=9.1\times 10^{-31}\ kg$

<em>Now, form the mathematical expression of Kinetic Energy:</em>

$KE= \frac{1}{2} m.v^2$

$1.92\times 10^{-19}=0.5\times 9.1\times 10^{-31}\times v^2$

$v^2=4.2198\times 10^{11}$

$v=6.496\times 10^6\ m.s^{-1}$

<u>from the relation of magnetic and centripetal forces we have the radius as:</u>

$r=\frac{m.v}{q.B}$

$r=\frac{9.1\times 10^{-31}\times 6.496\times 10^6 }{1.6\times 10^{-19}\times 0.07}$

$r=5.278\times 10^{-4}\ m$

6 0

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

Where do you feel that you are traveling at the fastest speed when on the swing?

il63 [147K]

Answer:

Explanation:

I think it's C, because at that point, you are going fastest. Sorry if im wrong, hope this helps.

7 0

3 years ago

Read 2 more answers

Luis and Aisha conducted an experiment. They exerted different forces on four objects. Their results are shown in the table.

lesya692 [45]

Answer:

Object 3 has greatest acceleration.

Explanation:

Objects Mass Force

1 10 kg 4 N

2 100 grams 20 N

3 10 grams 4 N

4 1 kg 20 N

Acceleration of object 1,

$a_1=\dfrac{F_1}{m_1}\\\\a_1=\dfrac{4}{10}\\\\a_1=0.4\ m/s^2$

Acceleration of object 2,

$a_2=\dfrac{F_2}{m_2}\\\\a_2=\dfrac{20}{0.1}\\\\a_2=200\ m/s^2$

Acceleration of object 3,

$a_3=\dfrac{F_3}{m_3}\\\\a_3=\dfrac{4}{0.01}\\\\a_3=400\ m/s^2$

Acceleration of object 4,

$a_4=\dfrac{F_4}{m_4}\\\\a_4=\dfrac{20}{1}\\\\a_3=20\ m/s^2$

It is clear that the acceleration of object 3 is $400\ m/s^2$ and it is greatest of all. So, the correct option is (3).

4 0

3 years ago

Read 2 more answers