Answer to Question #113370 in Engineering for Raphy

Question #113370
1. What gauge pressure in Pa must a pump produce in order to pump water from the bottom of Grand Canyon (730 m elevation) to the Indian gardens (1370 m)?

2. Calculate the pressure at the bottom of the container with 1.2 m high layer of oil (SG=0.82), 0.5 m layer of water and 0.3 m layer of mercury (SG=13.6)

3. A cube having a side 8 cm in length has a density of 900 kg/m³. What is the mass and unit weight of the cube?

4. Georgie is trying to float his toy boat on water during one rainy day. If it weighs 130 N and has a density of 9 g/cm³, what is the total volume of the toy and the buoyant force exerted on it?
1
Expert's answer
2020-05-05T03:47:08-0400

Solution.

1.h1=730m;h_1=730m;

h2=1370m;h_2=1370m;

p=ρgΔh;Δh=h2h1;p=\rho g\Delta h; \Delta h=h_2-h_1;

Δh=1370m730m=640m;\Delta h=1370m-730m=640m;

p=1000kg/m39.8N/kg640m=6.272106Pa;p=1000kg/m^3\sdot9.8N/kg\sdot640m=6.272\sdot10^6Pa;

2.ho=1.2m;ρo=0.82g/cm3=820kg/m3;h_o=1.2m; \rho_o=0.82g/cm^3=820kg/m^3;

hw=0.5m;ρw=1000kg/m3;h_w=0.5m; \rho_w=1000kg/m^3;

hm=0.3m;ρw=13.6g/cm3=13600kg/m3;h_m=0.3m; \rho_w=13.6g/cm^3=13600kg/m^3;

p=ρgh;p=\rho gh;

po=820kg/m39.8N/kg1.2m=9643.2Pa=9.643kPa;p_o=820kg/m^3\sdot 9.8N/kg\sdot1.2m=9643.2Pa=9.643kPa;

pw=1000kg/m39.8N/kg0.5m=4900Pa=4.9kPa;p_w=1000kg/m^3\sdot9.8N/kg\sdot0.5m=4900Pa=4.9kPa;

pm=13600kg/m39.8N/kg0.3m=39984Pa=39.984kPa;p_m=13600kg/m^3\sdot9.8N/kg\sdot0.3m=39984Pa=39.984kPa;

p=po+pw+pm;p=p_o+p_w+p_m;

p=9.643kPa+4.9kPa+39.984kPa=54.527kPa;p=9.643kPa+4.9kPa+39.984kPa=54.527kPa;

3.a=8cm=0.08m;a=8cm=0.08m;

ρ=900kg/m3;\rho=900kg/m^3;

m=ρV;m=\rho\sdot V;

V=a3;V=a^3; V=(0.08m)3=0.000512m3;V=(0.08m)^3=0.000512m^3;

m=900kg/m30.000512m3=0.4608kg;m=900kg/m^3\sdot0.000512m^3=0.4608kg;

P=mg;P=0.4608kg9.8N/kg=4.52N;P=mg; P=0.4608kg\sdot9.8N/kg=4.52N;

4.P=130N;P=130N;

ρ=9g/cm3=9000kg/m3;\rho=9g/cm^3=9000kg/m^3;

m=ρV    V=mρ;m=\rho V \implies V=\dfrac{m}{\rho};

P=mg    m=Pg;P=mg\implies m=\dfrac{P}{g};

m=130N9.8N/kg=13.265kg;m=\dfrac{130N}{9.8N/kg}=13.265kg;

V=13.265kg9000kg/m3=0.00147m3;V=\dfrac{13.265kg}{9000kg/m^3}=0.00147m^3;

FA=ρwgV;F_A=\rho_w gV;

FA=1000kg/m39.8N/kg0.00147m3=14.4N;F_A=1000kg/m^3\sdot9.8N/kg\sdot0.00147m^3=14.4N;

Answer: 1. p=6.272106Pa;p=6.272\sdot10^6Pa;

2.p=54.527kPa;p=54.527kPa;

3.m=0.4608kg;m=0.4608kg; P=4.52N;P=4.52N;

4.V=0.00147m3;V=0.00147m^3; FA=14.4N;F_A=14.4N;




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