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Vilka [71]
3 years ago
15

An alfisol is most closely related to ________.

Physics
1 answer:
Volgvan3 years ago
4 0
<span>Transitional environments</span>
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You are lying on a beach, your eyes 20cm above the sand. Just as the Sun sets, fully disappearing over the horizon, you immediat
Naya [18.7K]

Answer:

The  time is  t =  8.78 \ s

Explanation:

From the question we are told that  

  The  initial height of the eye  is  h _1 =  20 \  cm =  0.2 \  m

  The  height of the eye when you jumped up is h_2  =  150 \  cm  =  1.5 \  m

 An illustration of this question is shown on the first uploaded image

Generally the radius of the earth is  R  =  6.38*10^{6} \  m

Now from the diagram first sun means first time you saw the sun and  the second sun means second time you saw the sun then

   H is the height increase when you quickly stood up which is mathematically evaluated as

        H  =  h_2 -h_1

        H  = 1.5 -  0.2

        H  = 1.3 \ m

Also \theta i the angular displacement between the first and second position and from geometry it is also the angle at  one of the sides of the right angle triangle

Applying Pythagoras theorem

     (R+H)^2  =  K^2  + R^2

=>   R^2 + H^2 + 2RH  =  K^2  + R^2

Now given that H is very small compared to R the we ignore H^2

So

     R^2 + 2RH  =  K^2  + R^2

=>    K  =  \sqrt{2RH}

=>     K  =  \sqrt{2 *  6.38*10^6 * 1.3 }

=>      K  = 4073 \  m

Now the \theta is mathematically evaluated using SOHCAHTOA as follows

      tan  \theta  =  \frac{K}{R}

      \theta  = tan^{-1}[ \frac{K}{R}]

=>   \theta  = tan^{-1}[ \frac{ 4073}{6.38*10^{6}}]

=>   \theta = 0.0366^o

Generally

  1 \revolution\  around \ the\ earth =  24 \ hours =  86400 \ seconds =  360 ^o

So  

    \frac{\theta}{360}  =  \frac{t}{86400}

=>  \frac{0.0366}{360} =  \frac{t}{86400}

=>  t =  8.78 \ s

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Explanation:

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At t = 0, object A is dropped from the roof of a building. At the same instant, object B is dropped from a window 10 m below the
dsp73

Answer:

Explanation:

Given two objects are dropped simultaneously

Object A is 10 m higher than object B therefore

Distance covered by object A is given by

y_a(t) is given by

y=ut+\frac{1}{2}at^2

where y=displacement

u=initial velocity

a=acceleration

t=time

y_a(t)=0+0.5gt^2--1

for object B

y_b(t)=0+0.5gt^2--2

Subtract 1 and 2 we get

y_a(t)-y_b(t)=0

i.e. they will travel equal distance in equal time and distance between them remain 10 m until object B hits the ground

           

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