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3 years ago

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Looking down on a Northern Hemisphere extratropical cyclone, surface winds blow ________ about the center. a.counterclockwise an

d inward b.clockwise and outward

1 answer:

Lesechka [4]3 years ago

4 0

Answer:

counterclockwise and inward

Explanation:

Extratropical cyclones or mid-latitude cyclones are the low-pressure areas, along with anticyclones of the high-pressure areas that drive weather over much of the Earth.

<u>Due to the Coriolis effect, the flow of the wind around the extratropical cyclone is counterclockwise in direction in the northern hemisphere, and clockwise in direction in the southern hemisphere.</u>

<u>It forms a comma shape which drives the forces inward the center.</u>

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The bright yellow light emitted by a sodium vapor lamp consists of two emission lines at 589.0 and 589.6 nm. What are the freque

stiv31 [10]

Answer:

Explanation:

Given

wavelength of emissions are

$\lambda _1=589 nm$

$\lambda _2=589.6 nm$

Energy is given by

$E=\frac{hc}{\lambda }$

where h=Planck's constant

x=velocity of Light

$\lambda$ =wavelength of emission

$E_1=\frac{6.626\times 10^{-34}\times 3\times 10^8}{589\times 10^{-9}}$

$E_1=3.374\times 10^{-19} J$

$E_1 in kJ/mol$

$E_1=203.2 kJ/mol$

frequency corresponding to this emission

$\nu =\frac{c}{\lambda }$

$\nu _1=\frac{3\times 10^8}{589\times 10^{-9}}$

$\nu _1=5.09\times 10^{14} Hz$

Energy corresponding to $\lambda _2$

$E_2=\frac{6.626\times 10^{-34}\times 3\times 10^8}{589.6\times 10^{-9}}$

$E_2=3.371\times 10^{-19} J$

$E_2=203.02 kJ/mol$

frequency corresponding to this emission

$\nu =\frac{c}{\lambda }$

$\nu _1=\frac{3\times 10^8}{589.6\times 10^{-9}}$

$\nu _1=5.088\times 10^{14} Hz$

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3 years ago

As the spaceship travels upward in the sky, some of its kinetic energy will be lost to the universe due to ?

GaryK [48]

Answer:

Friction !!!

Explanation:

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3 years ago

What are the 2 types of ions.

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3 years ago

Which refers to an object’s resistance to any change in its motion? force acceleration gravity inertia

satela [25.4K]

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3 years ago

Read 2 more answers

(a) what is the acceleration of two falling sky divers (mass 132 kg including parachute) when the upward force of air resistance

Komok [63]

As per the question the mass of two falling sky drivers is 132 kg.

First we have to calculate their acceleration.

Whenever a body falls freely under gravity,its acceleration is acceleration due to gravity i.e g whose value is 9.8 m/s^2.

The earth pulls the object with a force equal to the weight of the body.

Hence the force gravity F=W= mg [ here m is mass of the body]

Here m =132 kg.

Hence force of gravity F= mg

=132 kg ×9.8 m/s^2

=1293.6 kg m/s^2

=1293.6 N [ here N[newton] is the unit of force.]

As per the question the air resistance is one fourth of weight of the bodies.

Hence air resistance F' =1/4 mg

$=\frac{1}{4} *1293.6N$

$=323.4 N$

Here F acts in vertically downward direction while F' acts in vertically upward direction.

Hence the net force acting on the particle is F-F'.

$F_{net} =1293.6N -323.4N$

$=970.2 N$

From Newton's second law of motion we know that net force is the product of mass and acceleration i.e

$F_{net} =ma$ [Here a is the acceleration]

$a =\frac{F_{net} }{m}$

$= \frac{970.2}{132} m/s^2$

$=7.35 m/s^2$

In the second question it has been told that they descend with uniform speed.hence acceleration of the two bodies will be zero.

we know that F= ma

=m×0

=0 N

Hence they will not get any force when they will descend with a uniform speed.

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3 years ago