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)

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)
by

1 Answer

Answer :

4474.23 N
by

Related questions

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

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

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 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 force required to accelerate amass of 30kg at a rate of 15m/s².  a. 460 N  b. 380 N  c. 560 N  d. 450 N F = ma

Description : What is the force required to accelerate amass of 30kg at a rate of 15m/s².  a. 460 N  b. 380 N  c. 560 N  d. 450 N F = ma

Last Answer : 450 N

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

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

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

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

Calculate the power output in horsepower of an 80-kg man that climbs a flight of stairs 3.8 m high in 4.0 s.  a) 744.8 hp  b) 0.998 hp  c) 746 hp  d) 1.998 hp Formula: Power = Fd/t = mgh/t F = W = mg d = h

Description : Calculate the power output in horsepower of an 80-kg man that climbs a flight of stairs 3.8 m high in 4.0 s.  a) 744.8 hp  b) 0.998 hp  c) 746 hp  d) 1.998 hp Formula: Power = Fd/t = mgh/t F = W = mg d = h

Last Answer : 0.998 hp

How many Newton’s (N) in 900,000 dynes?  a) 8 Newton’s  b) 9 Newton’s  c) 7 Newton’s  d) 6 Newton’s Formula: 1Newton (N)=100,000dynes

Description : How many Newton’s (N) in 900,000 dynes?  a) 8 Newton’s  b) 9 Newton’s  c) 7 Newton’s  d) 6 Newton’s Formula: 1Newton (N)=100,000dynes

Last Answer : 9 Newton’s

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

Archimedes principle states that  a. when a body is inversed in a fluid, the fluid exerts a downward force on the body whose magnitude is equal to the weight of the displaced fluid  b. when a body is inversed in a fluid, the fluid exerts a upward force on the body whose magnitude is equal to the weight of the displaced fluid.  c. when a body is inversed in a fluid, the fluid exerts a downward force on the body whose magnitude is greater than the weight of the displaced fluid.  d. when a body is inversed in a fluid, the fluid exerts a downward force on the body whose magnitude is lower than the weight of the displaced fluid.

Description : Archimedes principle states that  a. when a body is inversed in a fluid, the fluid exerts a downward force on the body whose magnitude is equal to the weight of the displaced fluid  b. ... fluid exerts a downward force on the body whose magnitude is lower than the weight of the displaced fluid.

Last Answer : when a body is inversed in a fluid, the fluid exerts a upward force on the body whose magnitude is equal to the weight of the displaced fluid.

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 1-kg steam-water mixture at 1.0 MPa is contained in an inflexible tank. Heat is added until the pressure rises to 3.5 MPa and the temperature to 400°. Determine the heat added.  a) 1378.7 kJ  b) 1348.5 kJ  c) 1278,7 kJ  d) 1246,5 kJ Formula: Q = (h2 – p2v2) –(h1 –p1v1)

Description : A 1-kg steam-water mixture at 1.0 MPa is contained in an inflexible tank. Heat is added until the pressure rises to 3.5 MPa and the temperature to 400°. Determine the heat added.  a) 1378.7 kJ  b) 1348.5 kJ  c) 1278,7 kJ  d) 1246,5 kJ Formula: Q = (h2 – p2v2) –(h1 –p1v1)

Last Answer : 1378.7 kJ

From the steam table, determine the average constant pressure specific heat (c) of steam at 10 kPa and45.8 ˚C  A.1.79 kJ/ kg-˚C  B.10.28 kJ/ kg-˚C  C.30.57 kJ/ kg-˚C  D. 100.1 kJ/ kg-˚C Formula: h = c T ∆ ∆ From the steam table At 47.7 ˚C h= 2588.1 kJ/ kg At 43.8 ˚C h= 2581.1 kJ/ kg

Description : From the steam table, determine the average constant pressure specific heat (c) of steam at 10 kPa and45.8 ˚C  A.1.79 kJ/ kg-˚C  B.10.28 kJ/ kg-˚C  C.30.57 kJ/ kg-˚C  D. 100.1 kJ/ kg-˚C Formula: h = c T ∆ ∆ From the steam table At 47.7 ˚C h= 2588.1 kJ/ kg At 43.8 ˚C h= 2581.1 kJ/ kg

Last Answer : 1.79 kJ/ kg-˚C

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

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

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

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

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

A vertical column of water will be supported to what height by standard atmospheric pressure. If the Y w = 62.4lb/ft3 po = 14.7 psi.  a. 44.9 ft  b. 33.9 ft  c. 22.9 ft  d. 55.9 ft formula: ho= po/Yw

Description : A vertical column of water will be supported to what height by standard atmospheric pressure. If the Y w = 62.4lb/ft3 po = 14.7 psi.  a. 44.9 ft  b. 33.9 ft  c. 22.9 ft  d. 55.9 ft formula: ho= po/Yw

Last Answer : 33.9 ft

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

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

Twenty grams of oxygen gas are compressed at a constant temperature of 30 ˚C to 5%of their original volume. What work is done on the system.  A.824 cal  B.924 cal  C.944 cal  D.1124 cal Formula: W = -mRTln (V2/V1) Where R = (1.98 cal/gmole·K) (32 g/gmole)

Description : Twenty grams of oxygen gas are compressed at a constant temperature of 30 ˚C to 5%of their original volume. What work is done on the system.  A.824 cal  B.924 cal  C.944 cal  D.1124 cal Formula: W = -mRTln (V2/V1) Where R = (1.98 cal/gmole·K) (32 g/gmole)

Last Answer : 1124 cal

Find the pressure due to a column of mercury 74.0 cm high.  a. 91.80 x 10^3 N/m^2  b. 73.56 x 10^2 N/m^2  c. 9.86 x 10^4 N/m^2  d. 87.2 x 10^4 N/m^2

Description : Find the pressure due to a column of mercury 74.0 cm high.  a. 91.80 x 10^3 N/m^2  b. 73.56 x 10^2 N/m^2  c. 9.86 x 10^4 N/m^2  d. 87.2 x 10^4 N/m^2

Last Answer : 9.86 x 10^4 N/m^2

The value of the product of molecular weight and the gas characteristic constant for all the gases in M.K.S. unit is  (a) 29.27 kgfm/mol°K  (b) 8314kgfm/mol°K  (c) 848kgfm/mol°K  (d) 427kgfm/mol°K  (e) 735 kgfm/mol°K.

Description : The value of the product of molecular weight and the gas characteristic constant for all the gases in M.K.S. unit is  (a) 29.27 kgfm/mol°K  (b) 8314kgfm/mol°K  (c) 848kgfm/mol°K  (d) 427kgfm/mol°K  (e) 735 kgfm/mol°K.

Last Answer : Answer : c

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

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

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

Description : 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

Last Answer : 4875 j

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 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)}

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

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

How much does an object having the mass of 100kg weight in newton.  a. 981 N  b. 991 N  c. 981.6 N  d. 980.1N F = ma

Description : How much does an object having the mass of 100kg weight in newton.  a. 981 N  b. 991 N  c. 981.6 N  d. 980.1N F = ma

Last Answer : 900,000lb/hr;625 ft/s981 N

What is the formula to convert °F to °C?  a) °C = °F + 273  b) °C = 5/9 (°F - 32)  c) °C = 9/5 (°F)+32  d) None of the above

Description : What is the formula to convert °F to °C?  a) °C = °F + 273  b) °C = 5/9 (°F - 32)  c) °C = 9/5 (°F)+32  d) None of the above

Last Answer : °C = 5/9 (°F - 32)

What is the formula to convert °C to °F?  a) °F = °C + 273  b) °F = 5/9 (°C - 32)  c) °F = 9/5 (°C)+32  d) None of the above

Description : What is the formula to convert °C to °F?  a) °F = °C + 273  b) °F = 5/9 (°C - 32)  c) °F = 9/5 (°C)+32  d) None of the above

Last Answer : momentum

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

The SI unit of mass  a. kg  b. g  c. N  d. None of the above

Description : The SI unit of mass  a. kg  b. g  c. N  d. None of the above

Last Answer : kg

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

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

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

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

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

What horse power is required to isothermally compress 800 ft^3 of Air per minute from 14.7 psia to 120 psia?  A. 28 hp  B.108 hp  C.256 hp  D.13900 hp Formula: W= p1V1 ln (p1/p2) Power = dW / dt

Description : What horse power is required to isothermally compress 800 ft^3 of Air per minute from 14.7 psia to 120 psia?  A. 28 hp  B.108 hp  C.256 hp  D.13900 hp Formula: W= p1V1 ln (p1/p2) Power = dW / dt

Last Answer : 108 hp

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

Calculate the entropy of steam at 60psiawith a quality of 0.8  A. 0.4274 BTU/lbm-˚R  B. 0.7303 BTU/lbm-˚R  C. 1.1577 BTU/lbm-˚R  D. 1.2172 BTU/lbm-˚R Formula: fromthe steamtable at 60 psia: sƒ = 0.4274 BTU/lbm-˚R sƒg = 1.2172 BTU/lbm-˚R) s = sƒ + x sƒg where x = is the quality

Description : Calculate the entropy of steam at 60psiawith a quality of 0.8  A. 0.4274 BTU/lbm-˚R  B. 0.7303 BTU/lbm-˚R  C. 1.1577 BTU/lbm-˚R  D. 1.2172 BTU/lbm-˚R Formula: fromthe steamtable at 60 psia: sƒ = 0.4274 BTU/lbm-˚R sƒg = 1.2172 BTU/lbm-˚R) s = sƒ + x sƒg where x = is the quality

Last Answer : 1.1577 BTU/lbm-˚R

To displace a cubic foot of fresh water, you need _________ force.  a. 62.4 lb  b. 9.81 lb  c. 76 lb  d. 760 lb

Description : To displace a cubic foot of fresh water, you need _________ force.  a. 62.4 lb  b. 9.81 lb  c. 76 lb  d. 760 lb

Last Answer : 62.4 lb

An ideal gas at 45psig and 80ºF is heated in the close container to 130ºF. What is the final pressure?  a. 65.10 psi  b. 65.11 psi  c. 65.23 psi  d. 61.16 psi P1V1/T1= P2V2/T2;V = Constant

Description : An ideal gas at 45psig and 80ºF is heated in the close container to 130ºF. What is the final pressure?  a. 65.10 psi  b. 65.11 psi  c. 65.23 psi  d. 61.16 psi P1V1/T1= P2V2/T2;V = Constant

Last Answer : 65.23 psi

Which one is the correct relation between energy efficiency ratio (EER) and coefficient of performance (COP)?  A. EER = 2.34 COP  B. EER = 3.24 COP  C. EER = 3.42 COP  D. EER = 4.23 COP

Description : Which one is the correct relation between energy efficiency ratio (EER) and coefficient of performance (COP)?  A. EER = 2.34 COP  B. EER = 3.24 COP  C. EER = 3.42 COP  D. EER = 4.23 COP

Last Answer : EER = 3.42 COP