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kolezko [41]
3 years ago
11

A storm front moves in and Rachel and Pam notice the column of mercury in the barometer rises only to 736 mm. Assume the density

of mercury is 13, 000 kg/m 3
(a) What is the change in air pressure?
(b) What if their barometer was filled with water instead of mercury, how high does the column rise? Density of water = 1000 kg/m
Physics
1 answer:
Orlov [11]3 years ago
6 0

Answer:

a

 \Delta  P =   7558.6 \  Pa

b

 h_1 =  10  \  m

Explanation:

From the question we are told that

   The position of the column of mercury in the barometer is  h = 736 \  mm = 0.76 \  m\

   The density of mercury is  \rho =  13,000 \  kg / m^3

Generally the  pressure of the atmosphere at that column is mathematically represented as  

         P  =  \rho  *  g * h

=>      P  =13 000   *  9.8  * 0.736

=>      P  = 93766.4 \  Pa

Generally the atmospheric pressure at sea level (Generally the pressure before the change in level of the mercury column)  is  P_a =  101325  \  Pa

Generally the change in air pressure  is mathematically represented as

      \Delta  P =  P_a - P

=>   \Delta  P =    101325  - 93766.4

=>   \Delta  P =   7558.6 \  Pa

Generally the height which the column will rise to is mathematically evaluated as  

         h_1 =  \frac{P}{ \rho_w *  g  }

Here \rho_w is the density of water with value  \rho_w =  1000 \ kg/m^3

So

       h_1 =  \frac{ 93766.4}{  1000 *  9.8   }

=>   h_1 =  10  \  m

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