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Nady [450]
3 years ago
8

3) Movement of water through the oceanic benthic zone.

Physics
1 answer:
Lubov Fominskaja [6]3 years ago
6 0

Answer:

I don't know the answer, but i hope you find it!! good luck

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A paper airplane has an acceleration of 5 m/s^2. If it is thrown from rest (0 m/s), how fast will it be going in 3 seconds?
FinnZ [79.3K]

Answer: its 0 i guess

Explanation:

7 0
3 years ago
Help Please fast like now!!​
defon

Answer: d. availability of places to play

Explanation: I think that it’s d because availability of space to do your physical activity is important. Availabilty of places to play is an environmental factor due to where you decide to play.

7 0
3 years ago
Earth orbits the sun once every
Karo-lina-s [1.5K]

Answer:1 trip around the earth is an angular displacement of 2*pi

3.6525*10^2 days

I

Explanation:24 h/1 day * 3.600*10^3 s/1h = 3.156*10^7 s

Angular speed = angular displacement / time

Angular speed = 2*pi rads / 3.156*10^7 s = 1.9910*10^-7 rad/s

3 0
3 years ago
Calculate the vapor pressure for a mist of spherical water droplets of radius 3.70×10−8m surrounded by water vapor at 298 K. The
poizon [28]

Answer:

The vapor pressure for a mist is P= 25.92\ Torr

Explanation:

From the question we are given that

        The radius is  r = 3.70 *10^{-8} m

        The temperature is T  = 298K

       The vapor pressure of water P_o = 25.2\ Torr

      The density of water is  \rho = 998 kg.m^{-3}

      The surface tension of water is \sigma  = 71.99 m N \cdot m^{-1}

Generally the equation of that is mathematically represented as

                            ln (\frac{P}{P_0} )  = \frac{2 \sigma M}{r \rho RT}

 Where  P is the vapor pressure for mist

               R is the  ideal gas constant = 8.31

      making P the subject in the formula

    P = e^ {\frac{2 \sigma M}{r \rho RT}} * P_0

        = e^{\frac{2 *(0.07199)(0.018)}{(3.70*10^{-8})(998)(8.31)(298)} } * 25.2

        P= 25.92Torr

               

4 0
3 years ago
Earth is about 150 million kilometers from the Sun, and the apparent brightness of the Sun in our sky is about 1300 watts/m2. Us
EleoNora [17]

3) Earth is about 150 million km from the Sun, and the apparent brightness of the Sun in our sky is about 1,300 watts per square meter. Determine the apparent brightness we would measure for the Sun if we were located five times Earth's distance from the Sun. Answer: The Sun would appear 1/25 times as bright.
4 0
4 years ago
Read 2 more answers
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