Answer all 8 Problems step by step (solution is provided): Answer for Number 2:

Question

Answer all 8 Problems step by step (solution is provided):

Answer for Number 2:

PA= 160 kPa

PB= 199 kPa

PC= 160 kPa

PD= 140 kPa

A vacuum suction wall hook is pressed against the wall, reducing the pressure in the interior 0.500 atm. The diameter of the hook is about 4.00 cm. Calculate the force required to separate it off the wall, 1 atm = 1.01 times 105 Pa (Answer: 63.4 N OR 14.2 pounds) Find the absolute hydrostatic water pressure at A, B, C and D shown in the diagram. All the valves are closed so water flow through the pipe is zero What is the total pressure p2 at the bottom of the tank?. The density of oil = 900 kg/m3 and the density of water = 1000 kg/m3. (1.23 times 105 Pa) What is the gas pressure inside the box shown below? (88000 Pa or 0.87 atm.) A 10 cm times 10cm times 10 cm wood block with a density of 700 kg/m3 floats in water. What is the distance from the tip of the block to the water if the water is fresh? (d = 3 cm) If it's seawater? (d = 3.2 cm) What is the tension in the string in figure shown below? (1.9 N) A 10 cm times 10cm times 10 cm block of steel (density = 7900 kg/m3) is suspended from a spring scale. The scale is in N. what is the scale reading if the block is in air? (77 N) What is the scale reading after the block has been lowered into a beaker of oil and is completely submerged? (69 N) A .spherical steel hall of radius 0.05 m is suspended with the help of a string as shown. (Consider the density of the steel 7800 kg/m3 The density of water is 1000 kg/m3 Use V (4/3)piR3 for the volume of a sphere where R is the radius of the sphere. Find the tension in the string: If the ball is in the air as shown in the figure (a) (40N) If the ball is completely immersed into the water as shown in the figure (b) (35N) If half of the volume of the ball is dipped into the water as shown in the figure (c) (37N)

Explanation / Answer

1. P= F/a

so force = P.a= (1.01x10^5 pa/2) 3.14*(0.02)^2= 63.4 N

2) PA= Patm + rho*g*hA = 1.01*10^5 pa +1000 *9.8*6 =159800 pa =160 kPa

PB= Patm + rho*g*hB =1.01*10^5 pa +1000 *9.8*10=199 kPa

PC= Patm + rho*g*hC=1.01*10^5 pa +1000 *9.8*6=160kPa

PD= Patm + rho*g*hD=1.01*10^5 pa +1000 *9.8*4=140 kPa

3) pressure at bottom = rho(oil)g*h1 +rho(water)*g*h2 = 900*9.8*(0.5) +1000*9.8*(1.2) =1.23*10^5 Pa

4)pressure in box = 1.01*10^5+(13600*9.8*0.06)-(13600*9.8*0.16)=87672 Pa

5) byouncay force = mg

volume submerged * density of liquid *g = m g

mass of block = 700 kg/m3 *(0.1*0.1*0.1)= 0.7 kg

for pure water

=>(0.1 *0.1*d)*1000 *9.8 = 0.7 *9.8

=>d = 3 cm

for sea water=density of sea water = 1035 kg/m3

=>(0.1 *0.1*d)*1000 *9.8 = 0.7 *9.8

=>d = 3.2 cm

6)density of ethyl alcohol = 789 kg/m3

mg - byouncy force=tension force

(0.0001m3)*789 kg/m3*9.8(0.0001*2700)*9.8 =1.9 N

7)when b;ock is in air

tension force = mg=(0.1*0.1*0.1)m3*7900 kg/m3*9.8m/s2 =77.42 N

b) density of oil is unknown here ..you can solve from byouncy force and mg balance

8a)tension in string = mg=(5.23*10^-4 *7800)*9.8= 40 N

when fully submerged

tension in string = mg - (Vol. submerged*density of liquid*g)

=(5.23*10^-4*7800*9.8)-(5.23*10^-4*1000*9.8)=34.85 N

when half submerged= mg - (Vol. submerged*density of liquid*g)

=(2.16*10^-4*7800*9.8)-(2.16*10^-4*1000*9.8)= 37 N

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