Tension Force
What is the tension in the string in the figure?
T = W - FB where T is the weightof the Al and FB buoyant force due to thealcohol
T = ρAl V g - ρE V g =(ρAl - ρE) V g =(ρAl - ρE) * 10E-4 *9.8 converting the volume to m3
Using 2700 for the density of Al and 790 for the density ofethyl alcohol
T = 1910 * 10E-4 * 9.8 = 1.87 N
A wood block with a density of 700 floats in water.
What is the distance from the top of the block to the water if the water is fresh? (a) density of fresh water is ρ_water = 1000 kg/m³
ΣF = 0
- (ρ_wood)g(V_wood) + (ρ_water)g(v_wood) = 0 (v_wood)/(V_wood) = (ρ_wood)/(ρ_water) (10 * 10 * h)/(10 * 10 * 10) = (700)/(1000) h = 7 cm the distance from the top of the block to the water is 10 - 7 = 3 cm (b) density of seawater is ρ_sea = 1030 kg/m³
(v_wood)/(V_wood) = (ρ_wood)/(ρ_sea)
(10 * 10 * h')/(10 * 10 * 10) = (700)/(1030) h' = 6.79611 cm the distance from the top of the block to the water is 10 - 6.79611 = 3.20389 cm
The container shown in the figure is filled with oil. It is open to the atmosphere on the left.
By the known formula, presure = P = h dg h is the level of height of the water column d is the density of the oil g is the acceleration due to gravity (a) presure at A : P_A = h d g = (0.5m -0m ) (900 kg/m^3)(9.8m/s^2)= 4410 Pa (b) presure between B and A = P_B - P_A = hdg = (0.5 m - 0 m )(900)(9.8)= 4410 Pa
(c) presure betwewn C and A = P_C - P_A = hdg = (0.5 m - 0 m )(900)(9.8)= 4410 Pa
What is the gas pressure inside the box shown in the figure?
Height difference between gas and mercury h = 16 cm –6 cm = 10 cm = 0.1 m P = Po + Dgh Where Po= atmospheric pressure = 1 atm = 101.3 * 10 ^ 3Pa D = mercury = 13600 kg / m^ 3Plug the values weget P = ( 101.3 * 10^ 3 Pa) +
(13328 Pa) = 114628 Pa (101325 Pa - 13328 Pa)=8.8*10^4
A 1.2-m-diameter vat of liquid is 2.4 deep. The pressure at the bottom of the vat is1.2 .
What
T = W - FB where T is the weightof the Al and FB buoyant force due to thealcohol
T = ρAl V g - ρE V g =(ρAl - ρE) V g =(ρAl - ρE) * 10E-4 *9.8 converting the volume to m3
Using 2700 for the density of Al and 790 for the density ofethyl alcohol
T = 1910 * 10E-4 * 9.8 = 1.87 N
A wood block with a density of 700 floats in water.
What is the distance from the top of the block to the water if the water is fresh? (a) density of fresh water is ρ_water = 1000 kg/m³
ΣF = 0
- (ρ_wood)g(V_wood) + (ρ_water)g(v_wood) = 0 (v_wood)/(V_wood) = (ρ_wood)/(ρ_water) (10 * 10 * h)/(10 * 10 * 10) = (700)/(1000) h = 7 cm the distance from the top of the block to the water is 10 - 7 = 3 cm (b) density of seawater is ρ_sea = 1030 kg/m³
(v_wood)/(V_wood) = (ρ_wood)/(ρ_sea)
(10 * 10 * h')/(10 * 10 * 10) = (700)/(1030) h' = 6.79611 cm the distance from the top of the block to the water is 10 - 6.79611 = 3.20389 cm
The container shown in the figure is filled with oil. It is open to the atmosphere on the left.
By the known formula, presure = P = h dg h is the level of height of the water column d is the density of the oil g is the acceleration due to gravity (a) presure at A : P_A = h d g = (0.5m -0m ) (900 kg/m^3)(9.8m/s^2)= 4410 Pa (b) presure between B and A = P_B - P_A = hdg = (0.5 m - 0 m )(900)(9.8)= 4410 Pa
(c) presure betwewn C and A = P_C - P_A = hdg = (0.5 m - 0 m )(900)(9.8)= 4410 Pa
What is the gas pressure inside the box shown in the figure?
Height difference between gas and mercury h = 16 cm –6 cm = 10 cm = 0.1 m P = Po + Dgh Where Po= atmospheric pressure = 1 atm = 101.3 * 10 ^ 3Pa D = mercury = 13600 kg / m^ 3Plug the values weget P = ( 101.3 * 10^ 3 Pa) +
(13328 Pa) = 114628 Pa (101325 Pa - 13328 Pa)=8.8*10^4
A 1.2-m-diameter vat of liquid is 2.4 deep. The pressure at the bottom of the vat is1.2 .
What