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

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
by

1 Answer

Answer :

42.41 ft^3
by

Related questions

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

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50'. What number is missing? -Riddles

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Last Answer : 22

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

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

A certain gas, with cp = 0.529Btu/ lb. °Rand R = 96.2ft.lb/lb. °R, expands from 5 cu ft and 80°F to 15 cu ft while the pressure remains constant at 15.5psia. Compute for T2.  a.1520°R  b. 1620°R  c. 1720°R  d. 1820°R formula: T2= T1V2/V1

Description : A certain gas, with cp = 0.529Btu/ lb. °Rand R = 96.2ft.lb/lb. °R, expands from 5 cu ft and 80°F to 15 cu ft while the pressure remains constant at 15.5psia. Compute for T2.  a.1520°R  b. 1620°R  c. 1720°R  d. 1820°R formula: T2= T1V2/V1

Last Answer : 1620°R

A certain gas, with cp = 0.529Btu/lb.°R and R = 96.2 ft.lb/lb.°R, expands from 5 cu ft and 80°F to 15 cu ft while the pressure remains constant at 15.5 psia. Compute for T2. (Formula: T2= T1V2/V1)  a. 460°R  b. 270°R  c. 1620 °R  d. None of the above

Description : A certain gas, with cp = 0.529Btu/lb.°R and R = 96.2 ft.lb/lb.°R, expands from 5 cu ft and 80°F to 15 cu ft while the pressure remains constant at 15.5 psia. Compute for T2. (Formula: T2= T1V2/V1)  a. 460°R  b. 270°R  c. 1620 °R  d. None of the above

Last Answer : 1620 °R

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

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

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

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

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Description : Calculate: a. Mass flow rate in lb/hr. b. The velocity at section 2 in fps  a. 800,000lb/hr;625ft/s  b. 900,000lb/hr;625 ft/s  c. 888,000lb/hr;269 ft/s  d. 700,000lb/hr;269 ft/s m = A1V!/V1

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

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

Ten cu. ft of air at 300psia and 400°F is cooled to 140°F at constant volume. What is the transferred heat?  a.-120Btu  b. -220Btu  c.-320Btu  d. -420Btu formula: Q= mcv(T2-T1)

Description : Ten cu. ft of air at 300psia and 400°F is cooled to 140°F at constant volume. What is the transferred heat?  a.-120Btu  b. -220Btu  c.-320Btu  d. -420Btu formula: Q= mcv(T2-T1)

Last Answer : -420Btu

Ten cu ft. of air at 300 psia 400°F is cooled to 140°F at constant volume. What is the final pressure? (formula: p2 = p1T2/T1)  a. 0  b. 209 psia  c. - 420 psia  d. None of the above

Description : Ten cu ft. of air at 300 psia 400°F is cooled to 140°F at constant volume. What is the final pressure? (formula: p2 = p1T2/T1)  a. 0  b. 209 psia  c. - 420 psia  d. None of the above

Last Answer : 209 psia

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

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

Assuming compression is according to the Law PV = C, Calculate the initial volume of the gas at a pressure of 2 bars w/c will occupy a volume of 6m³ when it is compressed to a pressure of 42 Bars.  a) 130m³  b) 136m³  c) 120m³  d) 126m³ Formula: P1V1/T1 =P2V2/T2

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Last Answer : 126m³

A fluid flows in a steady manner between two section in a flow line at section 1: A 1 = 1ft², V1 = 100fpm, volume1 of 4ft³/lb. at sec2: A2 = 2 ft², p= 0.20 lb/ft³ calculate the velocity at section 2.  a. 625 fpm  b. 567 fpm  c. 356 fpm  d. None of the above

Description : A fluid flows in a steady manner between two section in a flow line at section 1: A 1 = 1ft², V1 = 100fpm, volume1 of 4ft³/lb. at sec2: A2 = 2 ft², p= 0.20 lb/ft³ calculate the velocity at section 2.  a. 625 fpm  b. 567 fpm  c. 356 fpm  d. None of the above

Last Answer : 625 fpm

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

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

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

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

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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

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

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

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Last Answer : Answer : a

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

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

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Last Answer : 108 hp

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

In what ratio must a person mix three kinds of Oats costing him Rs 1.20,Rs 1.44 and Rs 1.74 per Kg so that the mixture may be worth Rs 1.41 per Kg? A)11:77:7 B) 11:45:7 C)25:45:8 D) 27:23:6

Description : In what ratio must a person mix three kinds of Oats costing him Rs 1.20,Rs 1.44 and Rs 1.74 per Kg so that the mixture may be worth Rs 1.41 per Kg? A)11:77:7 B) 11:45:7 C)25:45:8 D) 27:23:6

Last Answer : A)11:77:7

Water is flowing through a 1 foot diameter pipe at the rate of 10ft/sec. What is the volume flow rate of water in ft³/sec?  a. 7.85  b. 6.85  c. 8.85  d. 5.85 V = Aν

Description : Water is flowing through a 1 foot diameter pipe at the rate of 10ft/sec. What is the volume flow rate of water in ft³/sec?  a. 7.85  b. 6.85  c. 8.85  d. 5.85 V = Aν

Last Answer : 7.85

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

On volume basis, air contains following parts of oxygen  (a) 21  (b) 23  (c) 25  (d) 77  (e) 79.

Description : On volume basis, air contains following parts of oxygen  (a) 21  (b) 23  (c) 25  (d) 77  (e) 79.

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What is the resulting pressure when one pound of air at 15 psia and 200 ˚F is heated at constant volume to 800 ˚F?  A.15 psia  B. 28.6 psia  C. 36.4 psia.  D. 52.1 psia Formula : T1/p1 = T2/p2 p2= p1T2 / T1

Description : What is the resulting pressure when one pound of air at 15 psia and 200 ˚F is heated at constant volume to 800 ˚F?  A.15 psia  B. 28.6 psia  C. 36.4 psia.  D. 52.1 psia Formula : T1/p1 = T2/p2 p2= p1T2 / T1

Last Answer : 28.6 psia

The flow energy of 150 L of a fluid passing a boundary to a system is 110 kJ. Determine the pressure at this point  a. 733.33 kPa  b. 833.33 kPa  c. 933.33 kPa  d. 633.33 kPa

Description : The flow energy of 150 L of a fluid passing a boundary to a system is 110 kJ. Determine the pressure at this point  a. 733.33 kPa  b. 833.33 kPa  c. 933.33 kPa  d. 633.33 kPa

Last Answer : 733.33 kPa

An automobile weighing 2800 lb and travelling at 30 miles/hr, hits a depression in the road which has a radius of curvature of 50 ft. What is the total force to which springs are subjected. a.8230 lb b.6170 lb c.2800 ib d.107 dynes e.3280 lb

Description : An automobile weighing 2800 lb and travelling at 30 miles/hr, hits a depression in the road which has a radius of curvature of 50 ft. What is the total force to which springs are subjected. a.8230 lb b.6170 lb c.2800 ib d.107 dynes e.3280 lb

Last Answer : b. 6170 lb

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.

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

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

The ratio of two specific heats of air is equal to  (a) 0.17  (b) 0.24  (c) 0.1  (d) 1.41  (e) 2.71.

Description : The ratio of two specific heats of air is equal to  (a) 0.17  (b) 0.24  (c) 0.1  (d) 1.41  (e) 2.71.

Last Answer : Answer : d

Specific heat of air at constant pressure is equal to  (a) 0.17  (b) 0.21  (c) 0.24  (d) 1.0  (e) 1.41

Description : Specific heat of air at constant pressure is equal to  (a) 0.17  (b) 0.21  (c) 0.24  (d) 1.0  (e) 1.41

Last Answer : Answer : c

A certain fluid is flowing in a 0.5m x 0.3 channel at the rate of 3 m/s and has a specific volume of 0.012 m³/kg. Determined the mass of water flowing in kg/s.  a. 267 kg/s  b. 378 kg/s  c. 375 kg/s  d. 456.5 kg/s m = Aν/V

Description : A certain fluid is flowing in a 0.5m x 0.3 channel at the rate of 3 m/s and has a specific volume of 0.012 m³/kg. Determined the mass of water flowing in kg/s.  a. 267 kg/s  b. 378 kg/s  c. 375 kg/s  d. 456.5 kg/s m = Aν/V

Last Answer : 375 kg/s

What is defined as a process during which a fluid flows through a control volume steadily?  A. Transient-flow process  B. Steady and uniform process  C. Uniform-flow process  D. Steady-flow process

Description : What is defined as a process during which a fluid flows through a control volume steadily?  A. Transient-flow process  B. Steady and uniform process  C. Uniform-flow process  D. Steady-flow process

Last Answer : Steady-flow process

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

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 cantilever shaft having 50 mm diameter and a length of 300 mm has a disc of mass 100 kg at its free end. The Young’s modulus for the shaft material is 200 GN/m 2 . Determine the frequency of transverse vibrations of the shaft. a) 31 b) 35 c) 37 d) 41

Description : A cantilever shaft having 50 mm diameter and a length of 300 mm has a disc of mass 100 kg at its free end. The Young’s modulus for the shaft material is 200 GN/m 2 . Determine the frequency of transverse vibrations of the shaft. a) 31 b) 35 c) 37 d) 41

Last Answer : d) 41

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

The volume of a gas under standard atmospheric pressure & 76 cmHg is 200m³. What is the volume when pressure is 80 cmHg if the temperature is unchanged?  a) 180 in³  b) 170 in³  c) 160 in³  d) 190 in³ Formula: P2V2 = P1V1

Description : The volume of a gas under standard atmospheric pressure & 76 cmHg is 200m³. What is the volume when pressure is 80 cmHg if the temperature is unchanged?  a) 180 in³  b) 170 in³  c) 160 in³  d) 190 in³ Formula: P2V2 = P1V1

Last Answer : 190 in³

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