A 86871 lb force is applied to a surface that measures 94 in long and 58 in wide

Question

A 86871 lb force is applied to a surface that measures 94 in long and 58 in wide. Determine the pressure exerted on the surface by the force in both psi and Pa. 1 Pa = 1.45×10-4 psi.
___psi
___ Pa

2. Calculate how much pressure (due to the liquid) would be on a 100 kg person if they were 65 m deep in...

a.)...freshwater. The density of freshwater is 1000 kg/m3. Express your answer in both pascals and atmospheres. 1 atm = 1.013×105 Pa.
Pa
atm

b.)...saltwater. The density of saltwater is 1030 kg/m3. Express your answer in both pascals and atmospheres. 1 atm = 1.013×105 Pa.
Pa
atm

3. Determine the density of each object.

a.) Object 1: mass = 36.58 kg, volume = 3.5 L.
Density of Object 1 = kg/L

b.) Object 2: mass = 38.08 kg, volume = 2.8 L
Density of Object 2 = kg/L

c.) Object 3: mass = 41.38 kg, volume = 2.0 L
Density of Object 3 = kg/L

d.) Object 4: mass = 25024 kg, volume = 3680 L.
Density of Object 4 = kg/L

e.) Object 5: mass = 3298.68 kg, volume = 242.55 L.
Density of Object 5 = kg/L

Each of the five objects is to be completely submerged in mercury. The density of mercury is 13.6 kg/L.

f.) Which ones will rise? Choose all that apply.
Object 1
Object 2
Object 3
Object 4
Object 5

g.) Which ones will sink? Choose all that apply.
Object 1
Object 2
Object 3
Object 4
Object 5
A 81 kg rock with a volume of 17 L is completely submerged in water. The density of water is 1 kg/L.

a.) What volume of water is displaced by the rock?
L

b.) What mass of water is displaced by the rock?
kg

c.) What is the buoyant force on the rock?
N

5. Calculate how much a 4.2 m slab of steel will expand if heated from 6 °C to 242 °C. The coefficient of thermal expansion for steel is 11×10-6 . Express your answer in m, cm, & in. 1 in = 2.54 cm.
m
cm
in

6. Calculate how much heat is required to bring 292 g of water from a temperature of 4.2°C to a temperature of 100°C. The specific heat of water is 4186 J/kg·°C.
J
h.) Which ones will have neutral buoyancy? Choose all that apply.
Object 1
Object 2
Object 3
Object 4
Object 5

4.

Explanation / Answer

The force is 86871 lbf = 39404 kgf The area is A = (94 in)(58 in)                        = (2.387m)(1.473m) = 3.516 m^2 therefore the preesure               P = F/A                   = 39404*9.8 / 3.516                   = 1.1*10^5 Pa or 1.59 Psi (2) The pressure                    P = gh                       = (1000)(9.8)(65)                       = 6.37*10^5 Pa                      or 6.288 atm in salt water                    P = gh                       = (1030)(9.8)(65)                       = 6.56*10^5 Pa                      or 6.476 atm (3) The density of the object            = mass/volume               = 36.58/3.5L = 10.45 kg/L (b) The mass m = 38.08kg and volume v = 2.8L then the density = 38.08 / 2.8                               = 13.6 kg/L (c) The mass m = 41.38kg and volume V = 2.0L then denisty = 41.38/2 = 20.69 kg/L (d) m = 25024 kg and V = 3680L then density = 25024/3680 = 6.8 kg/L (e) m = 3298.68 kg and V = 242.55 L            then density = 3298.68/242.55                                  = 13.6 kg/L (5) If each the object submerged in density = 13.6 the density of the object is more than the liquid density will sink and which is less than the liquid is float above surface and which is equal is float the surface     the object (d) = 6.8 kg/L will rise the object (c) with density = 20.9kg/L will sink The mass of the rock m = 81kg the volume of the rock V = 17L (a) The volume of the water displaced v = 17L (b) The mass of the water displaed                 m = V = (1kg/L)(17) = 17kg (c) The buoyance force is equal to the mass of the water dispalced by object submerged             then F = 17kg*9.8 = 166.6 N                              = (1030)(9.8)(65)                       = 6.56*10^5 Pa                      or 6.476 atm (3) The density of the object            = mass/volume               = 36.58/3.5L = 10.45 kg/L (b) The mass m = 38.08kg and volume v = 2.8L then the density = 38.08 / 2.8                               = 13.6 kg/L (c) The mass m = 41.38kg and volume V = 2.0L then denisty = 41.38/2 = 20.69 kg/L (d) m = 25024 kg and V = 3680L then density = 25024/3680 = 6.8 kg/L (e) m = 3298.68 kg and V = 242.55 L            then density = 3298.68/242.55                                  = 13.6 kg/L (5) If each the object submerged in density = 13.6 the density of the object is more than the liquid density will sink and which is less than the liquid is float above surface and which is equal is float the surface     the object (d) = 6.8 kg/L will rise the object (c) with density = 20.9kg/L will sink The mass of the rock m = 81kg the volume of the rock V = 17L (a) The volume of the water displaced v = 17L (b) The mass of the water displaed                 m = V = (1kg/L)(17) = 17kg (c) The buoyance force is equal to the mass of the water dispalced by object submerged             then F = 17kg*9.8 = 166.6 N                                     

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