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4 years ago

What layer the atmosphere do you think is of greatest interest to meteorologists?

Physics

1 answer:

konstantin123 [22]4 years ago

3 0

The troposphere, because it's where the weather happens.

You might be interested in

Shown here are astronomical objects located at different distances from Earth. Rank the objects based on their distances from Ea

nevsk [136]

Answer: star on far side of Andromeda Galaxy- star on near side of Andromeda Galaxy- star on far side of Milky Way Galaxy- star near center of Milky Way Galaxy- Orion Nebula- Alpha Centauri- Pluto-The Sun

Explanation:

Andromeda is the nearest spiral galaxy to our Milky way galaxy. The star on the far side of the Andromeda galaxy would be farthest from Earth. Then, the star at the near side of the Andromeda galaxy.

The Earth along with the Sun and other planets are located in the Orion arm of the Milky way. It is about 8 kpc from the center of the Milky way. Thus, the next farthest astronomical object would be the star on the far side of the milky way and then a star near the center of the Milky way galaxy.

Orion Nebula is 1344 ly away from Earth. Alpha centauri is 4.36 ly away from earth. Pluto is about 28 AU away from Earth and Sun is 1 AU distance away.

4 0

3 years ago

Which of the following is the kinetic rate equation for the addition-elimination mechanism of nucleophilic aromatic substitution

Helga [31]

Answer:

Rate = k[aryl halide][nucleophile]

Explanation:

The simple aryl halides are almost inert to usual nucleophilic reagents but considerable activation on the ring can be produced by the addition of strongly electron-attracting substituents on either the ortho or para positions, or both. These groups deactivate the ring to allow the attack of the nucleophille on the ring.

Thus, these reactions can occur by following addition-elimination mechanism in which the nucleophille first attacks the aryl halide and then the elimination of the leaving group takes place.

Kinetic studies of this type of mechanism demonstrate that the reactions are of second-order kinetics– first order w.r.t. nucleophile and also, first-order w.r.t. aromatic substrate. The rate determining step (r.d.s.) is the formation of the addition intermediate.

Thus,

Rate = k[aryl halide][nucleophile]

7 0

3 years ago

A resonant circuit using a 286-nFnF capacitor is to resonate at 18.0 kHzkHz. The air-core inductor is to be a solenoid with clos

lukranit [14]

Answer:

The inductor contains $N = 523962.32$ loops

Explanation:

From the question we are told that

The capacitance of the capacitor is $C = 286nF = 286 * 10^{-9} \ F$

The resonance frequency is $f = 18.0 kHz = 18*10^{3} Hz$

The diameter is $d = 1.1 mm = \frac{1.1 }{1000} = 0.00011 \ m$

The of the air-core inductor is $l = 12 \ m$

The permeability of free space is $\mu_o = 4 \pi *10^{-7} \ T \cdot m/A$

Generally the inductance of this air-core inductor is mathematically represented as

$L = \frac{\mu_o * N^2 \pi d^2}{4 l}$

This inductance can also be mathematically represented as

$L = \frac{1}{w^2}$

Where $w$ is the angular speed mathematically given as

$w = 2 \pi f$

$L = \frac{1}{4 \pi ^2 f^2}$

Now equating the both formulas for inductance

$\frac{\mu_o * N^2 \pi d^2}{4 l} = \frac{1}{4 \pi ^2 f^2}$

making N the subject of the formula

$N = \sqrt{\frac{1}{(2 \pi f)^2} * \frac{4 * l }{\mu_o * \pi d^2 C} }$

$N = \frac{1}{2 \pi f} * \frac{2}{d} * \sqrt{\frac{l}{\pi * \mu_o * C} }$

Substituting value

$N = \frac{1}{ 3.142 * 18*10^{3} * 0.00011 } \sqrt{\frac{12}{ 3.142 * 4 \pi *10^{-7}* 286 *10^{-9}} }$

$N = 523962.32$ loops

4 0

3 years ago

Slick Willy is in traffic court (again) contesting a $50.00 ticket for running a red light. “You see, your Honor, as I was appro

natta225 [31]

Answer and Explanation:

$Greetings!$

$I'm~Isabelle~Williams~and~I~will~be~answering~your~question!$

$Given:$

$Wave~length~ of~ red ~ light = 683~ nm$

$Wave ~length ~of~ yellow~ light = 576~ nm$

$The~ speed ~of~ the~ Slick~ Willy~ can~ be:$

$\boxed{~f'=(\frac{v+v_{o} }{v} )f~}$

$Where:$

$f'=observed~frequency$

$f=actual~frequency$

$v=speed~of~light$

$v_{o}=~speed~of~observer$

$Plug~in~the~values:$

$683\times 10^-^9=576\times 10^-^9 ~(1+\frac{v_{o} }{3\times 10^8} )$

$\frac{683}{576}=1+\frac{v_{o} }{3\times 10^8}$

$\frac{v_{o} }{3\times 10^8} =\frac{683}{576}=-1$

$v_{o}=0.186\times 3\times 10^8$

$v_{o}=0.558\times 10^8 ~m/s$

$Hope~this~helps!~Have~an~amazing~day~ahead!~$

$-Isabelle~Williams$

6 0

4 years ago

•• Your roommate is working on his bicycle and has the bike upside down. He spins the 60-cm-diameter wheel, and you notice that

Y_Kistochka [10]

Answer:

$v=0.57\frac{m}{s}$

$a_c=10.83\frac{m}{s^2}$

Explanation:

We have an uniform circular motion, therefore, the pebble’s speed is given by the distance traveled in a revolution $(2\pi r)$ and the period (T), since this is the time pebble’s takes to complete a revolution: