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4 years ago

The atmosphere is composed of five?

2 answers:

6 0

The atmosphere is composed of five major layers. These layers are, in order from the surface of the earth, troposphere, stratosphere, mesosphere, ionosphere, and exosphere.

troposphere - lowest layer that provides majority of our weather. contains four-fifths of air on the earth and extends to a maximum height of 11 miles at the equator and lesser height at the poles.

stratosphere - extends to the height of 30 miles from the equator and it's the location of the ozone layer.

mesosphere - extends to the height of 52 miles from the equator and is where meteors generally burn up

ionosphere - extends to the height of 430 miles from the equator and is already considered as part of the outer space. It is where satellites orbit around the world.

exosphere - extends to the height range of 620 miles to 6,214 miles from the Earth's surface and merges with interplanetary space

motikmotik4 years ago

5 0

The atmosphere is made up of 5 major layers.

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3 years ago

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3 years ago

Alden, a passenger on a yacht moored 15 miles due north of a straight, east-west

Ghella [55]

Answer:

To minimize the travel time from the yacht to the hospital, the motorboat should head in a direction of 12.83 degrees west of south.

Explanation:

If we assume that both the motorboat and ambulance will be moving at a constant speed, we can calculate the time that each one will take to travel a given distance using the following equation:

$time=\frac{distance}{speed}$

Then the total travel time from the yacht to the hospital will be the motorboat travel time plus the ambulance travel time

$t=t_m+t_a$

$t=\frac{d_m}{s_m} +\frac{d_a}{s_a}$

First we must write the total travel time in terms of the motorboat's direction (Θ).

$cos(\theta)=\frac{15}{d_m}$

$d_m=\frac{15}{cos(\theta)}=15 sec(\theta)$

$d_a=60-d_1$

$tan(\theta)=\frac{d_1}{15}$

$d_1=15tan(\theta)$

$d_a=60-15tan(\theta)$

$t=t_m+t_a$

$t=\frac{d_m}{s_m} +\frac{d_a}{s_a}$

$t=\frac{15sec(\theta)}{20} + \frac{60-15tan(\theta)}{90}$

$t=\frac{15}{20}sec(\theta) + \frac{60}{90}-\frac{15}{90}tan(\theta)$

$t=\frac{3}{4}sec(\theta)-\frac{1}{6}tan(\theta) + \frac{2}{3}$

So this last equation represents the variation of the total travel time as a function of the motorboat's direction.

To find the equation's minimum point (which would be the direction with the minimum total travel time), we must find $\frac{dt}{d\theta}$ and then find its roots (its x-interceptions).

$\frac{dt}{d\theta}=\frac{d}{d\theta} (\frac{3}{4}sec(\theta)-\frac{1}{6}tan(\theta) + \frac{2}{3})$

$\frac{dt}{d\theta}=\frac{3}{4}sec(\theta)tan(\theta)-\frac{1}{6}sec^2(\theta)$

$\frac{dt}{d\theta}=sec(\theta)(\frac{3}{4}tan(\theta)-\frac{1}{6}sec(\theta))$

Now let's find the values of x which make $\frac{dt}{d\theta}=0$

$\frac{dt}{d\theta}=sec(\theta)(\frac{3}{4}tan(\theta)-\frac{1}{6}sec(\theta))=0$

As sec(\theta) is never equal to zero, then $\frac{dt}{d\theta}$ would be zero when

$\frac{3}{4}tan(\theta)=\frac{1}{6}sec(\theta)$

Graphing both equations we can find their interceptions and this would the value we're looking for.

In the attached images we can see that \theta=0.224 rad=12.83° is the minimum point for $t(\theta)$ . Then, to minimize the travel time from the yacht to the hospital, the motorboat should head in a direction of 12.83 degrees west of south.

6 0

3 years ago