Physics, asked by aayushbagri007, 8 months ago

calculate the depth below the surface of water at which the pressure due to the weight of water equals 1.00 atm.

Answers

Answered by deshdeepak88
4

We begin by solving the equation P =hρg for depth h :

h=P/ρg.

Then we take P to be 1.00 atm and ρ to be the density of the water that creates the pressure.

Solution:

Entering the known values into the expression for h gives

h=1.01×105N/m2(1.00×103kg/m3)(9.80m/s2)= 10.3m.

hope this will help you.

Attachments:
Answered by khadkaunish321
0

Answer:

The answer is 10.3m.

Explanation:

we know:

density of water(d)=1000kg/m^3

gravity(g)=9.8m/s^2

pressure(p)=1atm=1.01×10^5pa  ==>(given)

As we know pressure(p)=hdg,

then,

or,1.01×10^5=h×1000×9.8

or,h=1.01×10^5/(9.8×1000)

Therefore, h=10.3m.

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