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pychu [463]
3 years ago
12

Consider the double-fluid manometer attached to an air pipe shown. If the specific gravity of one fluid is 14.514.5, determine t

he specific gravity of the other fluid if pressure of air was 81kpa81kpa. Take the atmospheric pressure to be 100 kpa.

Physics
1 answer:
kvv77 [185]3 years ago
5 0

Complete Question

The complete question is shown on the first uploaded image  

Answer:

The  value  is  S_g_2  =  3.128

Explanation:

From the question we are told that  

  The specific gravity of the first fluid  is  S_g_1 =  14.5

  The  pressure of air is  P_a  =  81kPa  =  81*10^{3} \  Pa

  The  atmospheric pressure is  P_A  =  100 kPa  =  100 *10^{3}  Pa

Generally the specific gravity of the second  fluid is mathematically represented as

            S_g_2  =  S_g_1  * \frac{h_1}{ h_2 }  +  \frac{P_a  -  P_A}{\rho_w *  g *  h_2 }

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

=>   S_g_2  =  [14.5 *  \frac{0.22}{0.4} ] +  [\frac{81*10^{3}-100*10^{3}}{ 1000 *  9.8 *  0.4} ]

=> S_g_2  =  3.128

   

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Two small objects each with a net charge of +Q exert a force of magnitude F on each other. We replace one of the objects with an
Alona [7]

Answer:

F'= 4F/9

Explanation:

Two small objects each with a net charge of +Q exert a force of magnitude F on each other. If r is the distance between them, then the force is given by :

F=\dfrac{kQ^2}{r^2} ...(1)

Now, if one of the objects with another whose net charge is + 4Q is replaced and also the distance between +Q and +4Q charges is increased 3 times as far apart as they were. New force is given by :

F'=\dfrac{kQ\times 4Q}{(3r)^2}\\\\F'=\dfrac{4kQ^2}{9r^2}.....(2)

Dividing equation (1) and (2), we get :

\dfrac{F}{F'}=\dfrac{\dfrac{kQ^2}{r^2}}{\dfrac{4kQ^2}{9r^2}}\\\\\dfrac{F}{F'}=\dfrac{kQ^2}{r^2}\times \dfrac{9r^2}{4kQ^2}\\\\\dfrac{F}{F'}=\dfrac{9}{4}\\\\F'=\dfrac{4F}{9}

Hence, the correct option is (d) i.e. " 4F/9"

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a = F/m

when force F is kept constant , the acceleration is inversely related to the mass "m".  in other words, greater the mass , smaller will be the acceleration and smaller the mass, greater will be the acceleration.

the masses of baseball, basketball,  tennis ball and bowling ball can in arranged in increasing order as

tennis ball < baseball < basketball < bowling ball.

since acceleration and mass have inverse relation from the formula , the order of the acceleration will also be reverse.

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3 years ago
1 A 75-g ball is projected from a height of 1.6 m with a horizontal velocity of 2 m/s and bounces from a 400-g smooth plate supp
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Answer with explanation:

We are given that  

Mass of ball,m_1=75 g=\frac{75}{1000}=0075kg

1 kg=1000 g

Height,h_1=1.6 m

h_2=0.6 m

Horizontal velocity,v_x=2 m/s

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Where g=9.8 m/s^2

Velocity after impact,v_1=\sqrt{2gh_2}=\sqrt{2\times 9.8\times 0.6}=3.4m/s

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m_1u_1+m_2u_1=-m_1v_1+m_2v_2

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Final energy=\frac{1}{2}m_1v^2_x+m_1gh_2+\frac{1}{2}m_2v^2_2

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Energy lost due to compact=Initial energy-final energy=1.326-1.162=0.164 J

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