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AfilCa [17]
3 years ago
12

When transiting a great distance a Navigator prepares a __________ track so the ship can steady courses while driving the shorte

st distance.
Physics
2 answers:
coldgirl [10]3 years ago
7 0

Answer:

composite track

Explanation:

To travel an incredible circular track, the guide must constantly change course because the extraordinary circular track is a turn when plotted on a Mercator map. It is ridiculous to try to navigate an incredible circular route. All things considered, to make the best use of the shorter cruise separation from the extraordinary circular runway, pilots generally divide an incredible hover runway between the underlying position and the target into many much smaller sections ( for trajectory purposes) of approximately one to several days of cruising time (based clearly on the specialty and conditions) and making course changes every day simultaneously, generally in the early afternoon. Absolute separation is thus the set of separations of these fragments determined by the methods of Mercator Sailing. A potential problem with the incredible circular track, however, is the most limited route between two areas, similarly for most tracks closer to the well (or at a higher range) than the two points, starting point or goal. The high areas are often in danger due to the terrible climate and icing. A protected thought of a veteran sailor is to set a range limit for the long voyage plan. This arrangement is called an extraordinary composite circle course arrangement, terminated with way points. This minicomputer soothes the monotonous procedure for deciding these way points for travel.

Yakvenalex [24]3 years ago
6 0

Answer:

The answer is composite track.

Explanation:

For ships to cover the shortest distance between two points on the surface of the earth, navigators base their calculations using great circles. A great circle is circular line drawn on a globe that follows the circumference of the earth ( thereby dividing the globe into equal halves ).  As the ship moves from one point to the other, the navigator adjusts its course because the earth is on a constant rotation. Great circles, because they usually cover distances of about 40,000km are broken down into smaller lines called Rhumb so as to provide a steady course. The one common great circle is the equator and the ship's heading does not change on this line.

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Juan was wearing a bright red shirt in a very dark room. What color did his shirt appear to the people with him in the room? A)
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What kind of scattering (Rayleigh, Mie, or non-selective) would you expect to be most important when radiation of the specified
oee [108]

Answer:

hello your question is not properly arranged attached below is the arranged table and solution

answer : attached table below

Explanation:

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3 0
3 years ago
6) Find the speed a spherical raindrop would attain by falling from 4.00 km. Do this:a) In the absence of air dragb) In the pres
sleet_krkn [62]

We are asked to determine the velocity of a rain drop if it falls from 4 km.

To do that we will use the following formula:

2ah=v_f^2-v_0^2

Where:

\begin{gathered} a=\text{ acceleration} \\ h=\text{ height} \\ v_f,v_0=\text{ final and initial velocity} \end{gathered}

If we assume the initial velocity to be 0 we get:

2ah=v_f^2

The acceleration is the acceleration due to gravity:

2gh=v_f^2

Now, we take the square root to both sides:

\sqrt{2gh}=v_f

Now, we substitute the values:

\sqrt{2(9.8\frac{m}{s^2})(4000m)}=v_f

solving the operations:

280\frac{m}{s}=v

Therefore, the velocity without air drag is 280 m/s.

Part B. we are asked to determine the velocity if there is air drag. To do that we will use the following formula:

F_d=\frac{1}{2}C\rho_{air}Av^2

Where:

\begin{gathered} F_d=drag\text{ force} \\ C=\text{ constant} \\ \rho_{air}=\text{ density of air} \\ A=\text{ area} \\ v=\text{ velocity} \end{gathered}

We need to determine the drag force. To do that we will use the following free-body diagram:

Since the velocity that the raindrop reaches is the terminal velocity and its a constant velocity this means that the acceleration is zero and therefore the forces are balanced:

F_d=mg

Now, we determine the mass of the raindrop using the following formula:

m=\rho_{water}V

Where:

\begin{gathered} \rho_{water}=\text{ density of water} \\ V=\text{ volume} \end{gathered}

The volume is the volume of a sphere, therefore:

m=\rho_{water}(\frac{4}{3}\pi r^3)

Since the diameter of the raindrop is 3 millimeters, the radius is 1.5 mm or 0.0015 meters. Substituting we get:

m=(0.98\times10^3\frac{kg}{m^3})(\frac{4}{3}\pi(0.0015m)^3)

Solving the operations:

m=1.39\times10^{-5}kg

Now, we substitute the values in the formula for the drag force:

F_d=(1.39\times10^{-5}kg)(9.8\frac{m}{s^2})

Solving the operations:

F_d=1.36\times10^{-4}N

Now, we substitute in the formula:

1.36\times10^{-4}N=\frac{1}{2}C\rho_{air}Av^2

Now, we solve for the velocity:

\frac{1.36\times10^{-4}N}{\frac{1}{2}C\rho_{air}A}=v^2

Now, we substitute the values. We will use the area of a circle:

\frac{1.36\times10^{-4}N}{\frac{1}{2}(0.45)(1.21\frac{kg}{m^3})(\pi r^2)}=v^2

Substituting the radius:

\frac{1.36\cdot10^{-4}N}{\frac{1}{2}(0.45)(1.21\frac{kg}{m^{3}})(\pi(0.0015m)^2)}=v^2

Solving the operations:

70.67\frac{m^2}{s^2}=v^2

Now, we take the square root to both sides:

\begin{gathered} \sqrt{70.67\frac{m^2}{s^2}}=v \\  \\ 8.4\frac{m}{s}=v \\  \end{gathered}

Therefore, the velocity is 8.4 m/s

7 0
1 year ago
Question below...............
kondaur [170]

Answer:

friction force

Explanation:

force of friction is opposite to the force applied it resist the motion

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2 years ago
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