In the above problem, find the potential energy of the mass with respect to datum. (Formula: P = mgz/k )  a. 4875 j  b. 0.51 j  c. 0.46 j  d. None of the above

In the above problem, find the
potential energy of the mass with
respect to datum.
(Formula: P = mgz/k )
 a. 4875 j
 b. 0.51 j
 c. 0.46 j
 d. None of the above
by

1 Answer

Answer :

4875 j
by

Related questions

A mass of 5kg is 100m above a given datum where local g = 9.75 m/s2 . Find the gravitational force in newtons. (Formula: Fg= mg/k )  a. 48.75 N  b. 50 N  c. 45 N  d. None of the above

Description : A mass of 5kg is 100m above a given datum where local g = 9.75 m/s2 . Find the gravitational force in newtons. (Formula: Fg= mg/k )  a. 48.75 N  b. 50 N  c. 45 N  d. None of the above

Last Answer : 48.75 N

For a certain gas R = 320 J/kg.K and cv= 0.84kJ/kg.K. Find k?  a. 1.36  b. 1.37  c. 1.38  d. 1.39 formula: k= R / cv+1

Description : For a certain gas R = 320 J/kg.K and cv= 0.84kJ/kg.K. Find k?  a. 1.36  b. 1.37  c. 1.38  d. 1.39 formula: k= R / cv+1

Last Answer : 1.38

Utilizing the answer to the previous problem, estimate the overall or average increase in temperature ( ΔT) of the concrete roof from the energy absorbed from the sun during a12hour day. Assume that all of the radiation absorbed goes into heating the roof. The specific heat of concrete is about 900 J/kg, and the density is about 2,300 kg/m3 .  a. 7.9 °C  b. 8.9°C  c. 9.9°C  d. 10.9°C formula: ΔQ = m c ΔT

Description : Utilizing the answer to the previous problem, estimate the overall or average increase in temperature ( ΔT) of the concrete roof from the energy absorbed from the sun during a12hour day. Assume that all of the radiation absorbed goes into ... °C  b. 8.9°C  c. 9.9°C  d. 10.9°C formula: ΔQ = m c ΔT

Last Answer : 7.9 °C

A perfect gas has a value of R= 319.2 J/ kf.K and k= 1.26. If 120 kJ are added to 2.27 kf\g of this gas at constant pressure when the initial temp is 32.2°C? Find T2.  a. 339.4 K  b. 449.4 K  c. 559.4K  d. 669.4K formula: cp = kR/ k-1 Q= mcp(T2-T1)

Description : A perfect gas has a value of R= 319.2 J/ kf.K and k= 1.26. If 120 kJ are added to 2.27 kf\g of this gas at constant pressure when the initial temp is 32.2°C? Find T2.  a. 339.4 K  b. 449.4 K  c. 559.4K  d. 669.4K formula: cp = kR/ k-1 Q= mcp(T2-T1)

Last Answer : 339.4 K

There are 1.36 kg of gas, for which R = 377 J/kg.k and k = 1.25, that undergo a nonflow constant volume process from p1 = 551.6 kPa and t1 = 60°C to p2 = 1655 kPa. During the process the gas is internally stirred and there are also added 105.5 kJ of heat. Determine t2. (Formula: T2= T1p2/ p1)  a. 999 K  b. 888 K  c. 456 K  d. One of the above

Description : There are 1.36 kg of gas, for which R = 377 J/kg.k and k = 1.25, that undergo a nonflow constant volume process from p1 = 551.6 kPa and t1 = 60°C to p2 = 1655 kPa. During the process the gas is internally stirred and ... (Formula: T2= T1p2/ p1)  a. 999 K  b. 888 K  c. 456 K  d. One of the above

Last Answer : 999 K

Helium ( R= 0.4698 BTU/lbm-˚R ) is compressed isothermally from 14.7 psia and 68 ˚F. The compression ratio is 1:4. Calculate the work done by the gas.  A. –1454 BTU/lbm  B. -364 BTU/lbm  C.-187BTU/lbm  D.46.7 BTU/lbm Formula: W = RT ln (V2/V1)

Description : Helium ( R= 0.4698 BTU/lbm-˚R ) is compressed isothermally from 14.7 psia and 68 ˚F. The compression ratio is 1:4. Calculate the work done by the gas.  A. –1454 BTU/lbm  B. -364 BTU/lbm  C.-187BTU/lbm  D.46.7 BTU/lbm Formula: W = RT ln (V2/V1)

Last Answer : -364 BTU/lbm

The combined mass of car and passengers travelling at 72 km/hr is 1500 kg. Find the kinetic energy of this combined mass. (Formula: K =mv2 / 2k )  a. 300 kJ  b. 200 kJ  c. 500 kJ  d. None of the above

Description : The combined mass of car and passengers travelling at 72 km/hr is 1500 kg. Find the kinetic energy of this combined mass. (Formula: K =mv2 / 2k )  a. 300 kJ  b. 200 kJ  c. 500 kJ  d. None of the above

Last Answer : 300 kJ

A 0.064 kg of octane vapor (MW = 114) is mixed with0.91 kg of air (MW = 29.0) in the manifold of an Engine. The total pressure in the manifold is 86.1 kPa, and a temperature is 290 K. assume octane behaves ideally. What is the partial pressure of the air in the mixture in KPa?  a. 46.8  b. 48.6  c. 84.6  d. 64.8

Description : A 0.064 kg of octane vapor (MW = 114) is mixed with0.91 kg of air (MW = 29.0) in the manifold of an Engine. The total pressure in the manifold is 86.1 kPa, and a temperature is 290 K. assume octane behaves ideally. What is ... of the air in the mixture in KPa?  a. 46.8  b. 48.6  c. 84.6  d. 64.8

Last Answer : 84.6

In the above problem, compute for the mass. (Formula: m = p1V1 / RT1)  a. 0.2148 lb  b. 0.2134 lb  c. 0.1248 lb  d. None of the above

Description : In the above problem, compute for the mass. (Formula: m = p1V1 / RT1)  a. 0.2148 lb  b. 0.2134 lb  c. 0.1248 lb  d. None of the above

Last Answer : 0.2148 lb

Two masses, one of the 10kg and the other unknown, are placed on a scale in a region where g = 9.67 m/sec2 . The combined weight of these two masses is 174.06 N. Find the unknown mass in kg.  a. 20 kg  b. 19 kg  c. 18 kg  d. 17 kg formula: m=Fg k / g

Description : Two masses, one of the 10kg and the other unknown, are placed on a scale in a region where g = 9.67 m/sec2 . The combined weight of these two masses is 174.06 N. Find the unknown mass in kg.  a. 20 kg  b. 19 kg  c. 18 kg  d. 17 kg formula: m=Fg k / g

Last Answer : 18 kg

A car whose mass is 2 metric tons is accelerated uniformly from stand hill to 100 kmph in 5 sec. Find the driving force in Newton’s.  a. 11,120 N  b. 11,320 N  c. 11,420 N  d. 11520 N formula: F= ma / k

Description : A car whose mass is 2 metric tons is accelerated uniformly from stand hill to 100 kmph in 5 sec. Find the driving force in Newton’s.  a. 11,120 N  b. 11,320 N  c. 11,420 N  d. 11520 N formula: F= ma / k

Last Answer : 11,120N

What form of energy refers to those a system possesses as a whole with respect to some outside reference frame, such as potential and kinetic energies?  A. Macroscopic form of energy  B. Microscopic form of energy  C. Internal energy  D. External energy

Description : What form of energy refers to those a system possesses as a whole with respect to some outside reference frame, such as potential and kinetic energies?  A. Macroscopic form of energy  B. Microscopic form of energy  C. Internal energy  D. External energy

Last Answer : Macroscopic form of energy

Refer to problem # 11. Determine the force that accelerates if to 12 m/s^2. horizontally along frictionless plane.  A. 2474.23 N  B. 2574.23 N  C. 3474.23 N  D. 2374.23 N Formula : M = wk / g F = ma /k

Description : Refer to problem # 11. Determine the force that accelerates if to 12 m/s^2. horizontally along frictionless plane.  A. 2474.23 N  B. 2574.23 N  C. 3474.23 N  D. 2374.23 N Formula : M = wk / g F = ma /k

Last Answer : 2474.23 N

Two thick slices of bread, when completely oxidized by the body, can supply 200,000 cal of heat. How much work is this equivalent to?  a) 4,190,000 joules  b) 8,390,000 joules  c) 839,000 joules  d) 419 000 joules Formula: J =Work/Heat J = mechanical equivalent of heat whose value is 4.19 joules/calorie

Description : Two thick slices of bread, when completely oxidized by the body, can supply 200,000 cal of heat. How much work is this equivalent to?  a) 4,190,000 joules  b) 8,390,000 joules  c) 839, ... d) 419 000 joules Formula: J =Work/Heat J = mechanical equivalent of heat whose value is 4.19 joules/calorie

Last Answer : 419 000 joules

How many joules of work is the equivalent of 15000 cal of heat?  a) 62850 joules  b) 3579.95 joules  c) 14995.81 joules  d) 15004.19 joules Formula: J =Work/Heat J = mechanical equivalent of heat whose value is 4.19 joules/calorie

Description : How many joules of work is the equivalent of 15000 cal of heat?  a) 62850 joules  b) 3579.95 joules  c) 14995.81 joules  d) 15004.19 joules Formula: J =Work/Heat J = mechanical equivalent of heat whose value is 4.19 joules/calorie

Last Answer : 62850 joules

Find ∫ for steam at 100 psia and 600°F.If h = 1329.6 and v = 6.216  a. 1214 Btu / lb  b. 1234 Btu /lb  c. 1342 Btu / lb  d. 1324 Btu /lb formula: ∫ = h– pv/ J

Description : Find ∫ for steam at 100 psia and 600°F.If h = 1329.6 and v = 6.216  a. 1214 Btu / lb  b. 1234 Btu /lb  c. 1342 Btu / lb  d. 1324 Btu /lb formula: ∫ = h– pv/ J

Last Answer : 1214Btu / lb

The value for the ΔU of a system is -120 J. If the system is known to have absorbed 420 J of heat, how much work was done?  a. -540 J  b. -640 J  c. -740 J  d. -840 J formula: ΔU = q +w

Description : The value for the ΔU of a system is -120 J. If the system is known to have absorbed 420 J of heat, how much work was done?  a. -540 J  b. -640 J  c. -740 J  d. -840 J formula: ΔU = q +w

Last Answer : -540 J

The concrete roof of a house is 10 m by 8 m and 10 cm thick (4"). Estimate the total heat the roof would absorb over the 12 day?  a. 1.3 x 108 J  b 2.3 x 108 J  c. 3.3 x 108 J  d. 4.3 x 108 J formula: ΔQ = ΔQ/Δtx Δt

Description : The concrete roof of a house is 10 m by 8 m and 10 cm thick (4"). Estimate the total heat the roof would absorb over the 12 day?  a. 1.3 x 108 J  b 2.3 x 108 J  c. 3.3 x 108 J  d. 4.3 x 108 J formula: ΔQ = ΔQ/Δtx Δt

Last Answer : 1.3 x 108 J

Let a closed system execute a state change for which the heat is Q = 100 J and work is W = -25 J. Find E. ∆ (Formula: E = Q- W) ∆  a. 125 J  b. 123 J  c. 126 J  d. None of the above

Description : Let a closed system execute a state change for which the heat is Q = 100 J and work is W = -25 J. Find E. ∆ (Formula: E = Q- W) ∆  a. 125 J  b. 123 J  c. 126 J  d. None of the above

Last Answer : 125 J

Water (specific heat cv= 4.2 kJ/ kg ∙ K ) is being heated by a 1500 W h eater. What is the rate of change in temperature of 1kg of the water?  A. 0.043 K/s  B. 0.179 K/s  C. 0.357 K/s  D. 1.50 K/s Formula: Q = mcv ( T)

Description : Water (specific heat cv= 4.2 kJ/ kg ∙ K ) is being heated by a 1500 W h eater. What is the rate of change in temperature of 1kg of the water?  A. 0.043 K/s  B. 0.179 K/s  C. 0.357 K/s  D. 1.50 K/s Formula: Q = mcv ( T)

Last Answer : 0.179 K/s

Twenty grams of ice at 0˚C melts to water at 0˚C. How much does the entropy of the 20g change in this process?  a. 30.5 J/K  b. 24.6 J/K  c. 21.3 J/K  d. 15.7 J/K

Description : Twenty grams of ice at 0˚C melts to water at 0˚C. How much does the entropy of the 20g change in this process?  a. 30.5 J/K  b. 24.6 J/K  c. 21.3 J/K  d. 15.7 J/K

Last Answer : 24.6 J/K

Steam at 1000 lbf/ft^2 pressure and 300˚R has specific volume of 6.5 ft^3/lbm and a specific enthalpy of 9800 lbf-ft/lbm. Find the internal energy per pound mass of steam.  A.2500 lbf-ft/lbm  B.3300 lbf-ft/lbm  C.5400 lbf-ft/lbm  D.6900 lbf-ft/lbm Formula: h= u+ pV u= h– pV

Description : Steam at 1000 lbf/ft^2 pressure and 300˚R has specific volume of 6.5 ft^3/lbm and a specific enthalpy of 9800 lbf-ft/lbm. Find the internal energy per pound mass of steam.  A.2500 lbf-ft/lbm  B.3300 lbf-ft/lbm  C.5400 lbf-ft/lbm  D.6900 lbf-ft/lbm Formula: h= u+ pV u= h– pV

Last Answer : 3300 lbf-ft/lbm

Gas is enclosed in a cylinder with a weighted piston as the stop boundary. The gas is heated and expands from a volume of 0.04 m^3 to 0.10 m^3 at a constant pressure of 200kPa.Calculate the work done by the system.  A. 8 kJ  B. 10 kJ  C.12 kJ  D.14 kJ Formula: W = p(V2-V1)

Description : Gas is enclosed in a cylinder with a weighted piston as the stop boundary. The gas is heated and expands from a volume of 0.04 m^3 to 0.10 m^3 at a constant pressure of 200kPa.Calculate the work done by the system.  A. 8 kJ  B. 10 kJ  C.12 kJ  D.14 kJ Formula: W = p(V2-V1)

Last Answer : 12 kJ

3.0 lbm of air are contained at 25 psia and 100 ˚F. Given that Rair = 53.35 ft-lbf/lbm- ˚F, what is the volume of the container?  A.10.7 ft^3  B.14.7 ft^3  C.15 ft^3  D.24.9 ft^3 Formula: use the ideal gas law pV = mRT T = (100 +460) ˚R V = mRT/p

Description : 3.0 lbm of air are contained at 25 psia and 100 ˚F. Given that Rair = 53.35 ft-lbf/lbm- ˚F, what is the volume of the container?  A.10.7 ft^3  B.14.7 ft^3  C.15 ft^3  D.24.9 ft^3 Formula: use the ideal gas law pV = mRT T = (100 +460) ˚R V = mRT/p

Last Answer : 24.9 ft^3

A vessel with a volume of cubic meter contains liquid water and water vapor ion equilibrium at 600 kPa. The liquid water has mass of1kg. Using the steam table, calculate the mass of the water vapor.  A. 0.99kg  B. 1.57 kg  C. 2.54 kg  D. 3.16 kg Formula: from the steam table at 600kPa vƒ = 0.001101 m^3 / kg vg = 0.3157 m^3 / kg Vtot = mƒ vƒ + mg vg mg = (tot-mƒ vƒ) / vg

Description : A vessel with a volume of cubic meter contains liquid water and water vapor ion equilibrium at 600 kPa. The liquid water has mass of1kg. Using the steam table, calculate the mass of the water vapor.  A. 0.99kg  B. 1.57 kg  C ... / kg vg = 0.3157 m^3 / kg Vtot = mƒ vƒ + mg vg mg = (tot-mƒ vƒ) / vg

Last Answer : 3.16 kg

A 30-m vertical column of fluid (density 1878 kg/m3 ) is located where g= 9.65 mps2 . Find the pressure at the base of the column. (Formula: pg= gρhg/k )  a. 543680 N/m2  b. 543.68 kPa (gauge)  c. Both a & b  d. None of the above

Description : A 30-m vertical column of fluid (density 1878 kg/m3 ) is located where g= 9.65 mps2 . Find the pressure at the base of the column. (Formula: pg= gρhg/k )  a. 543680 N/m2  b. 543.68 kPa (gauge)  c. Both a & b  d. None of the above

Last Answer : Both a & b

What is the specific weight of water at standard condition? (Formula: γ = ρg / k)  a. 1000 kgm/m3  b. 9.8066 m/s2  c. 1000 kgf/m3  d. None of the above

Description : What is the specific weight of water at standard condition? (Formula: γ = ρg / k)  a. 1000 kgm/m3  b. 9.8066 m/s2  c. 1000 kgf/m3  d. None of the above

Last Answer : 1000 kgf/m3

What is the weight of a66-kgm man at standard condition? (Formula: Fg= mg / k)  a. 66 kgf  b. 66 kgm  c. 66 lbm  d. 66 gf

Description : What is the weight of a66-kgm man at standard condition? (Formula: Fg= mg / k)  a. 66 kgf  b. 66 kgm  c. 66 lbm  d. 66 gf

Last Answer : 66 kgf

A system weighing 2kN. Determine the force that accelerate if to 12 m/s^2. a. vertically upward when g = 9.7 m/s^2  A. 4474.23 N  B.5484.23 N  C.4495.23 N  D.5488.23 N Formula: F = m/k (a +g)

Description : A system weighing 2kN. Determine the force that accelerate if to 12 m/s^2. a. vertically upward when g = 9.7 m/s^2  A. 4474.23 N  B.5484.23 N  C.4495.23 N  D.5488.23 N Formula: F = m/k (a +g)

Last Answer : 4474.23 N

A vertical column of water will be supported to what height by standard atmospheric pressure.  a. 33.9 ft  b. 45 ft  c. 67 ft  d. 25.46 ft ho= Po/Yo

Description : A vertical column of water will be supported to what height by standard atmospheric pressure.  a. 33.9 ft  b. 45 ft  c. 67 ft  d. 25.46 ft ho= Po/Yo

Last Answer : 33.9 ft

While swimming at depth of120m in a fresh water lake, A fish emits an air bubbles of volume 2.0mm³ atmospheric pressure is 100kPa. What is the pressure of the bubble?  a) 217.7 kPa  b) 317.7 kPa  c) 417.7 kPa  d) 517.7 kPa Formula: P= δh

Description : While swimming at depth of120m in a fresh water lake, A fish emits an air bubbles of volume 2.0mm³ atmospheric pressure is 100kPa. What is the pressure of the bubble?  a) 217.7 kPa  b) 317.7 kPa  c) 417.7 kPa  d) 517.7 kPa Formula: P= δh

Last Answer : 217.7 kPa

A pressure gage registers 50 psig in a region where the barometer is 14.25 psia. Find absolute pressure in psia, Pa. (Formula; p = patm+ pg)  a. 433 kPa  b. 443 kPa  c. 343 kPa  d. None of the above

Description : A pressure gage registers 50 psig in a region where the barometer is 14.25 psia. Find absolute pressure in psia, Pa. (Formula; p = patm+ pg)  a. 433 kPa  b. 443 kPa  c. 343 kPa  d. None of the above

Last Answer : 443 kPa

If air is at pressure, p, of 3200 lbf/ft2 , and at a temperature, T, of 800 ˚R, what is the specific volume, v? (R=5303 ft-lbf/lbm-˚R, and air can be modeled as an ideal gas.)  A.9.8 ft^3/lbm  B.11.2 ft^3/lbm  C.13.33 ft^3/lbm  D.14.2 ft^3/lbm Formula: pv = RT v = RT / p

Description : If air is at pressure, p, of 3200 lbf/ft2 , and at a temperature, T, of 800 ˚R, what is the specific volume, v? (R=5303 ft-lbf/lbm-˚R, and air can be modeled as an ideal gas.)  A.9.8 ft^3/lbm  B.11.2 ft^3/lbm  C.13.33 ft^3/lbm  D.14.2 ft^3/lbm Formula: pv = RT v = RT / p

Last Answer : 13.33 ft^3/lbm

Which of the following is the Ideal gas law (equation)?  A. V/T = K  B. V= k*(1/P)  C. P1/T1 = P2/T2  D. PV = nRT

Description : Which of the following is the Ideal gas law (equation)?  A. V/T = K  B. V= k*(1/P)  C. P1/T1 = P2/T2  D. PV = nRT

Last Answer : PV = nRT

What mass of nitrogen is contained in a10 ft3 vessel at a pressure of 840atm and 820°R? Make a computation by using ideal gas equation.  a. 194lb  b. 214lb  c. 394 lb  d. 413lb formula: m=pV /RT

Description : What mass of nitrogen is contained in a10 ft3 vessel at a pressure of 840atm and 820°R? Make a computation by using ideal gas equation.  a. 194lb  b. 214lb  c. 394 lb  d. 413lb formula: m=pV /RT

Last Answer : 394 lb

The weight of an object is 50lb. What is its mass at standard condition?  a. 50 lbm  b. 60 lbm  c. 70 lbm  d. 80 lbm formula: m = Fgk /g

Description : The weight of an object is 50lb. What is its mass at standard condition?  a. 50 lbm  b. 60 lbm  c. 70 lbm  d. 80 lbm formula: m = Fgk /g

Last Answer : 50 lbm

A problem Drum ( 3 ft. diameter ; 6 ft. height ) is field with a fluid whose density is 50 lb/ft^3. Determine the total volume of the fluid.  A. 42.41 ft^3  B.44.35 ft^3  C.45.63 ft^3  D.41.23 ft^3 Formula: Vf = (pi d^2 h) / 4

Description : A problem Drum ( 3 ft. diameter ; 6 ft. height ) is field with a fluid whose density is 50 lb/ft^3. Determine the total volume of the fluid.  A. 42.41 ft^3  B.44.35 ft^3  C.45.63 ft^3  D.41.23 ft^3 Formula: Vf = (pi d^2 h) / 4

Last Answer : 42.41 ft^3

A cylinder contains oxygen at a pressure of 10 atm and a temperature of 300 K. The volume of the cylinder is 10 liters. What is the mass of the oxygen in grams? Molecular weight (MW) of oxygen is 32 g/mole?  a. 125.02  b. 130.08  c. 135.05  d. 120.04

Description : A cylinder contains oxygen at a pressure of 10 atm and a temperature of 300 K. The volume of the cylinder is 10 liters. What is the mass of the oxygen in grams? Molecular weight (MW) of oxygen is 32 g/mole?  a. 125.02  b. 130.08  c. 135.05  d. 120.04

Last Answer : 130.08 {(10atm)(10)(32)/(0.0821) (300K)}

Find the change in internal energy of 5 lb. of oxygen gas when the temperature changes from 100 ˚F to 120 ˚F. CV = 0.157 BTU/lbm-˚R  A.14.7 BTU  B.15.7 BTU  C. 16.8 BTU  D. 15.9 BTU Formula: U= mcv T

Description : Find the change in internal energy of 5 lb. of oxygen gas when the temperature changes from 100 ˚F to 120 ˚F. CV = 0.157 BTU/lbm-˚R  A.14.7 BTU  B.15.7 BTU  C. 16.8 BTU  D. 15.9 BTU Formula: U= mcv T

Last Answer : 15.7 BTU

A 10m^3 vessel initially contains 5 m^3 of liquid water and 5 m^3 of saturated water vapor at 100 kPa. Calculate the internal energy of the system using the steam table.  A. 5 x10^5 kJ  B. 8x10^5 kJ  C. 1 x10^6 kJ  D. 2 x10^6 kJ Formula: fromthe steamtable vƒ = 0.001043 m^3 / kg vg = 1.6940 m^3 / kg u ƒ= 417.3 kJ/kg ug= 2506kJ/kg formula: Mvap = V vap/vg M liq = Vliq/ vƒ u =uƒM liq + ug M vap

Description : A 10m^3 vessel initially contains 5 m^3 of liquid water and 5 m^3 of saturated water vapor at 100 kPa. Calculate the internal energy of the system using the steam table.  A. 5 x10^5 kJ  B. 8x10^5 kJ  C. 1 ... 3 kJ/kg ug= 2506kJ/kg formula: Mvap = V vap/vg M liq = Vliq/ vƒ u =uƒM liq + ug M vap

Last Answer : 2 x10^6 kJ

If a system absorbs 500 cal of heat at the same time does 400J of work, find the change in internal energy of the system.  a. 1400 J  b. 1700 J  c. 1900 J  d. 1500 J

Description : If a system absorbs 500 cal of heat at the same time does 400J of work, find the change in internal energy of the system.  a. 1400 J  b. 1700 J  c. 1900 J  d. 1500 J

Last Answer : 1700 J

What is the amount of radiant energy received each second over each square meter that is at right angles to the sun’s rays at the top of the atmosphere?  a. 1400 J  b. 6000 J  c. 10000 J  d. 800 J

Description : What is the amount of radiant energy received each second over each square meter that is at right angles to the sun’s rays at the top of the atmosphere?  a. 1400 J  b. 6000 J  c. 10000 J  d. 800 J

Last Answer : 1400 J

Kinetic energy of the molecules in terms of absolute temperature (T) is proportional to  (a) T  (b) j  (c) J2  (d) Vr  (e) 1/Vr.

Description : Kinetic energy of the molecules in terms of absolute temperature (T) is proportional to  (a) T  (b) j  (c) J2  (d) Vr  (e) 1/Vr.

Last Answer : Answer : a

The value of gas constant (R) in S. I. units is  A. 0.287 J/kgK  B. 2.87 J/kgK  C. 28.7 J/kgK  D. 287 J/kgK

Description : The value of gas constant (R) in S. I. units is  A. 0.287 J/kgK  B. 2.87 J/kgK  C. 28.7 J/kgK  D. 287 J/kgK

Last Answer : Answer: D

_________ is the energy stored within a body or substance by virtue of the activity and configuration of its molecules.  a. Internal Energy  b. External Energy  c. Kinetic Energy  d. Potential Energy

Description : _________ is the energy stored within a body or substance by virtue of the activity and configuration of its molecules.  a. Internal Energy  b. External Energy  c. Kinetic Energy  d. Potential Energy

Last Answer : Internal Energy

The energy that stored in a system as a result of its position in the earth’s gravitational field  a. elastic energy  b. kinetic energy  c. potential energy  d. flow energy

Description : The energy that stored in a system as a result of its position in the earth’s gravitational field  a. elastic energy  b. kinetic energy  c. potential energy  d. flow energy

Last Answer : potential energy

A 600kg hammer of a pile driver is lilted 2m the pilling head. What is the change of potential energy? If the hammer is realest. What will be its velocity and the instant if it sticks the pilling?  a. 10,772 N-m and 5.26m/s  b. 13,200 N-m and 5.26m/s  c. 11,772 N-m and 6.26m/s  d. 11,77 2N-m and5.26m/s ∫PE = mgo(∫Z)/gc

Description : A 600kg hammer of a pile driver is lilted 2m the pilling head. What is the change of potential energy? If the hammer is realest. What will be its velocity and the instant if it sticks the pilling?  a. 10,772 N-m and 5.26m/s ...  c. 11,772 N-m and 6.26m/s  d. 11,77 2N-m and5.26m/s ∫PE = mgo(∫Z)/gc

Last Answer : 11,772 N-m and 6.26m/s

_________ is a thermodynamic potential that measures the “useful” or process-initiating work obtainable from an isothermal, isobaric thermodynamic system.  a. Du-Pont Potential  b. Gibbs free energy  c. Rabz-Eccles Energy  d. Claussius Energy

Description : _________ is a thermodynamic potential that measures the “useful” or process-initiating work obtainable from an isothermal, isobaric thermodynamic system.  a. Du-Pont Potential  b. Gibbs free energy  c. Rabz-Eccles Energy  d. Claussius Energy

Last Answer : Gibbs free energy

The energy of molecular motion appears as  (a) heat  (b) potential energy  (c) surface tension  (d) friction  (e) increase in pressure.

Description : The energy of molecular motion appears as  (a) heat  (b) potential energy  (c) surface tension  (d) friction  (e) increase in pressure.

Last Answer : Answer : a

The flow energy of 5 ft3 of a fluid passing a boundary to a system is 80,000 ft-lb. Determine the pressure at this point.  a. 222 psi  b. 333 psi  c. 444 psi  d. 111 psi formula: Ef= pV

Description : The flow energy of 5 ft3 of a fluid passing a boundary to a system is 80,000 ft-lb. Determine the pressure at this point.  a. 222 psi  b. 333 psi  c. 444 psi  d. 111 psi formula: Ef= pV

Last Answer : 111 psi