The best and most correct answer among the choices provided by the question is the first choice "warm, dry air"
In meteorology, precipitation<span> is any product of the condensation of atmospheric water vapor that falls under gravity. The main forms of </span>precipitation<span> include drizzle, rain, sleet, snow, graupel and hail.</span>
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Long straight distance that a person can swim is 5.64 m.
<h3>What is the
Long straight distance?</h3>
The line that runs form one end of the circle to another is called the diameter of the circle. The pool is a circle according to the question and the long straight distance that a person can swim is the same of the diameter of the circular pool.
Now we have;
A = πr^2
A = area of pool
r = radius of pool
r = √A/ π
r = √25/3.142
r = 2.82m
Diameter of the circular pool = 2 r = 2 (2.82 cm) = 5.64 m
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Missing parts;
An ad for an above-ground pool states that it is 25 m2. From the ad, you can tell that the pool is a circle. If you swim from one point at the edge of the pool to another, along a straight line, what is the longest distance d you can swim? Express your answer in three significant figures.
Given: A cubic tank holds 1,000.0 kg of water.
Mass of water in tank (m) = 1000.0 kg
Density of water (d) = 1000.0 kg /m³
Concept: Volume(V) = Mass / Density
Since the tank holds these water in it so the volume of water will be equal to the volume of the tank.
Hence, the volume of the tank = Mass of water / Density of water
or, = 1000.0 kg / 1000.0 kg m⁻³
or, = 1.0 m³
Since tank is cubical in shape. Let its side be 'x'
The volume of tank (x³) = 1.0 m³
or. side of tank (x) = 1.0 m
Hence, the dimensions of the tank will be 1.0 m.
Answer:
Explanation:
Newton's law of universal gravitation states that the force experimented by a satellite of mass m orbiting Mars, which has mass 
at a distance r will be:
where 
is the gravitational constant.
This force is the centripetal force the satellite experiments, so we can write:
Putting all together:
which means:
![r=\sqrt[3]{\frac{GM}{4\pi^2}T^2}](https://tex.z-dn.net/?f=r%3D%5Csqrt%5B3%5D%7B%5Cfrac%7BGM%7D%7B4%5Cpi%5E2%7DT%5E2%7D)
Which for our values is:
![r=\sqrt[3]{\frac{(6.67\times10^{-11}Nm^2/kg^2)(6.39\times10^{23} kg)}{4\pi^2}(1.026\times24\times60\times60s)^2}=20395282m=20395.3km](https://tex.z-dn.net/?f=r%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B%286.67%5Ctimes10%5E%7B-11%7DNm%5E2%2Fkg%5E2%29%286.39%5Ctimes10%5E%7B23%7D%20kg%29%7D%7B4%5Cpi%5E2%7D%281.026%5Ctimes24%5Ctimes60%5Ctimes60s%29%5E2%7D%3D20395282m%3D20395.3km)
Since this distance is measured from the center of Mars, to have the height above the Martian surface we need to substract the radius of Mars R=3389.5 km
, which leaves us with:
