Tuesday

October 13, 2015
Total # Posts: 781

**Math (help steve)**

Brilli the ant placed the numbers 1,2,…,n in order clockwise around a circle. She can create an infinite sequence A={aj}∞j=0 of integers by letting a0=k∈{1,2,…,n} and constructing ai+1 from ai by taking the integer that is ai positions clockwise in the ...
*July 16, 2013*

**Math (algebra)**

What are the last three digits of the largest perfect square that can be expressed as p^2+pq+q^2, where p and q are (positive) prime numbers?
*July 16, 2013*

**Math (algebra)**

Let x,y be complex numbers satisfying x+y=a xy=b, where a and b are positive integers from 1 to 100 inclusive. What is the sum of all possible distinct values of a such that x^3+y^3 is a positive prime number?
*July 16, 2013*

**Math (number theory)**

For how many positive integers N is ⌊N^2/5⌋(floor of N^2/5) a prime?
*July 16, 2013*

**Physics**

Telephoto lenses allow one to "zoom in" on distant objects, making them bigger. A particular telephoto lens consists of a combination of two thin lenses having focal lengths of f1=20 cm and f2=−8 cm, respectively. The lenses are separated by a distance of d=15 ...
*July 16, 2013*

**Physics Help**

1. The total enthalpy of 15.000kg of steam at 2250.000kPa is 34191.510kJ. Determine the dryness of the steam. ANS = % 2. Determine the quantity of heat required, to raise 11.100kg of water at 80.000 degrees Celsius, to saturated steam at 2250.000kPa and 71.400% dry. ANS = kJ ...
*July 14, 2013*

**Physics (Help)**

1. The total enthalpy of 15.000kg of steam at 2250.000kPa is 34191.510kJ. Determine the dryness of the steam. ANS = % 2. Determine the quantity of heat required, to raise 11.100kg of water at 80.000 degrees Celsius, to saturated steam at 2250.000kPa and 71.400% dry. ANS = kJ ...
*July 14, 2013*

**Thermodynamics **

Posted by Danny on Saturday, July 13, 2013 at 9:42am. 1. The total enthalpy of 15.000kg of steam at 2250.000kPa is 34191.510kJ. Determine the dryness of the steam. ANS = % 2. Determine the quantity of heat required, to raise 11.100kg of water at 80.000 degrees Celsius, to ...
*July 14, 2013*

**Thermodynamics**

1. The total enthalpy of 15.000kg of steam at 2250.000kPa is 34191.510kJ. Determine the dryness of the steam. ANS = % Not sure where to start. I need a little help please. I do have a chart with the properties of saturated steam. Thank you.
*July 13, 2013*

**Thermodynamics**

2. Determine the quantity of heat required, to raise 11.100kg of water at 80.000 degrees Celsius, to saturated steam at 2250.000kPa and 71.400% dry. ANS = kJ Not sure where to start. I need a little help please. I do have a chart with the properties of saturated steam. Thank ...
*July 13, 2013*

**Physics**

2. Determine the quantity of heat required, to raise 11.100kg of water at 80.000 degrees Celsius, to saturated steam at 2250.000kPa and 71.400% dry. ANS = kJ Not sure where to start. I need a little help please. I do have a chart with the properties of saturated steam. Thank ...
*July 13, 2013*

**Physics**

1. The total enthalpy of 15.000kg of steam at 2250.000kPa is 34191.510kJ. Determine the dryness of the steam. ANS = % Not sure where to start. I need a little help please. I do have a chart with the properties of saturated steam. Thank you.
*July 13, 2013*

**Thermodynamics**

1. The total enthalpy of 15.000kg of steam at 2250.000kPa is 34191.510kJ. Determine the dryness of the steam. ANS = % 2. Determine the quantity of heat required, to raise 11.100kg of water at 80.000 degrees Celsius, to saturated steam at 2250.000kPa and 71.400% dry. ANS = kJ ...
*July 13, 2013*

**Thermodynamics**

1. The total enthalpy of 15.000kg of steam at 2250.000kPa is 34191.510kJ. Determine the dryness of the steam. ANS = % 2. Determine the quantity of heat required, to raise 11.100kg of water at 80.000 degrees Celsius, to saturated steam at 2250.000kPa and 71.400% dry. ANS = kJ ...
*July 12, 2013*

**Physics(Thermodynamics)**

1. The total enthalpy of 15.000kg of steam at 2250.000kPa is 34191.510kJ. Determine the dryness of the steam. ANS = % 2. Determine the quantity of heat required, to raise 11.100kg of water at 80.000 degrees Celsius, to saturated steam at 2250.000kPa and 71.400% dry. ANS = kJ ...
*July 12, 2013*

**Thermodynamics **

1. The total enthalpy of 15.000kg of steam at 2250.000kPa is 34191.510kJ. Determine the dryness of the steam. ANS = % 2. Determine the quantity of heat required, to raise 11.100kg of water at 80.000 degrees Celsius, to saturated steam at 2250.000kPa and 71.400% dry. ANS = kJ ...
*July 11, 2013*

**Thermodynamics**

27.300kg of Copper at 35.000 degrees Celsius receives 460.000kJ of heat. Calculate the final temperature of the metal. ANS = degrees Celsius Copper specific heat = 0.39kJ/kg degrees Celsius. Q = mc (t2 - t1) Q = 460.000kJ m = 27.300kg c = 0.39kJ/kg t1 = 35 degrees Celsius t2...
*July 5, 2013*

**Thermodynamics**

Calculate the amount of expansion occurring when a steel rail 920.000 meters in length is heated from -10.000 to 410.000 degrees Celsius. ANS = m Steel = 12 * 10 ^-6 = 0.000012 Change in length = Original length * Coefficient of linear expansion * Temperature rise. = 920m * 0....
*July 4, 2013*

**Physics (Thermodynamics)**

27.300kg of Copper at 35.000 degrees Celsius receives 460.000kJ of heat. Calculate the final temperature of the metal. ANS = degrees Celsius Copper specific heat = 0.39kJ/kg degrees Celsius. Q = mc (t2 - t1) Q = 460.000kJ m = 27.300kg c = 0.39kJ/kg t1 = 35 degrees Celsius t2...
*July 4, 2013*

**Physics**

The Ultimate strength of a round steel rod is 630.000MPa. If a factor of safety of 5.000 is required, what is the maximum permissible load for the rod if it has a diameter of 3.000cm? ANS = kN (Round to 3 decimal places) Safety = 5.000 Rod diameter = 3.000cm = 0.03m Ultimate ...
*July 3, 2013*

**Physics**

What is the factor of safety of a steel hanger having an ultimate strength of 630.000MPa and supporting a load of 65000.000N. The steel hanger in question has a cross sectional area of 5.300cm. ANS = (Round to 3 decimal places) Load = 65000.000N Ultimate Strength = 630.000MPa...
*July 3, 2013*

**Physics**

Wouldn't you use Allowable Strength and not Ultimate Strength? Factor of Safety = Ultimate Strength/ Allowable Strength Allowable Strength = Ultimate Strength/ Factor of Safety Allowable Strength = Load/ Area= Load= Allowable Strength * Area= = Allowable Strength * (pi*d^2...
*July 3, 2013*

**Physics**

The Ultimate strength of a round steel rod is 630.000MPa. If a factor of safety of 5.000 is required, what is the maximum permissible load for the rod if it has a diameter of 3.000cm? ANS = kN (Round to 3 decimal places) Safety = 5.000 Rod diameter = 3.000cm = 0.03m Ultimate ...
*July 3, 2013*

**Physics**

A steel bar has a rectangular cross section measuring 1.000cm by 2.900cm and is 8.700m in length. A load of 189000.00N causes the bar to extend 1.500mm. Calculate the stress and strain produced. ANS 1 = kPa (3 decimal places) ANS 2 = (6 decimal places) 1.000cm = 0.01m 2.900cm...
*July 3, 2013*

**Physics**

The Cross Sectional Area is 10cm^2. Does that mean that the bar would be square in shape? 10cm = 0.10m 10cm^2 = 0.10m * 0.10m * 0.10m * 0.10m = 10 * 10^-4 = 0.001
*July 3, 2013*

**Physics**

A uniform bar 1.000 meters in length and having a cross sectional area of 10.000cm^2 is subjected to a loading of 5750.000 Newtons. This load causes an extension to the bar length of 2.740mm. Calculate the stress and the strain produced. ANS 1 = kPa (Round to 3 decimal places...
*July 3, 2013*

**Physics**

1) Calculate the tensile stress in a 33.000mm diameter rod subjected to a pull of 30.000kN. ANS= MPa (Round to 3 decimal places) Diameter = 33.000mm = 0.033m 30.000kN = 30000N Area = pi*d^2/4 = 3.1416 * 0.033^2/ 4 = 3.1416 *0.001089/ 4 = 0.0034212024/ 4 = 8.553006*10^-4 = 0....
*July 3, 2013*

**Physics**

The ultimate strength of a steel rod is 600.000MPa. if the factor of safety 3.000 is required, what is the maximum permissible load for the rod if it has a diameter of 6.000cm? ANS = kN (Round to 3 decimal place) Ultimate strength = 600.000MPa = 600000000Pa = 600 * 10^6 Factor...
*July 3, 2013*

**Physics**

Sorry..ANS = 148440.600kN not kPa. Please Check. Thank you
*July 2, 2013*

**Physics**

The Ultimate strength of a round steel rod is 630.000MPa. If a factor of safety of 5.000 is required, what is the maximum permissible load for the rod if it has a diameter of 3.000cm? ANS = kN (Round to 3 decimal places) Safety = 5.000 Rod diameter = 3.000cm = 0.03m Ultimate ...
*July 2, 2013*

**Physics**

A uniform bar 1.000 meters in length and having a cross sectional area of 10.000cm^2 is subjected to a loading of 5750.000 Newtons. This load causes an extension to the bar length of 2.740mm. Calculate the stress and the strain produced. ANS 1 = kPa (Round to 3 decimal places...
*July 2, 2013*

**Physics**

A steel bar has a rectangular cross section measuring 1.000cm by 2.900cm and is 8.700m in length. A load of 189000.00N causes the bar to extend 1.500mm. Calculate the stress and strain produced. ANS 1 = kPa (3 decimal places) ANS 2 = (6 decimal places) 1.000cm = 0.01m 2.900cm...
*July 2, 2013*

**Physics**

The Ultimate strength of a round steel rod is 630.000MPa. If a factor of safety of 5.000 is required, what is the maximum permissible load for the rod if it has a diameter of 3.000cm? ANS = kN (Round to 3 decimal places) Safety = 5.000 Rod diameter = 3.000cm = 0.03m Ultimate ...
*July 2, 2013*

**Physics**

What is the factor of safety of a steel hanger having an ultimate strength of 630.000MPa and supporting a load of 65000.000N. The steel hanger in question has a cross sectional area of 5.300cm. ANS = (Round to 3 decimal places) Load = 65000.000N ultimate strength = 630.000MPa...
*July 2, 2013*

**Physics**

What is the factor of safety of a steel hanger having an ultimate strength of 630.000MPa and supporting a load of 65000.000N. The steel hanger in question has a cross sectional area of 5.300cm. ANS = (Round to 3 decimal places) Load = 65000.000N ultimate strength = 630.000MPa...
*July 1, 2013*

**Physics**

The Ultimate strength of a round steel rod is 630.000MPa. If a factor of safety of 5.000 is required, what is the maximum permissible load for the rod if it has a diameter of 3.000cm? ANS = kN (Round to 3 decimal places) Safety = 5.000 Rod diameter = 3.000cm = 0.03m Ultimate ...
*July 1, 2013*

**Physics**

A steel bar has a rectangular cross section measuring 1.000cm by 2.900cm and is 8.700m in length. A load of 189000.00N causes the bar to extend 1.500mm. Calculate the stress and strain produced. ANS 1 = kPa (3 decimal places) ANS 2 = (6 decimal places) 1.000cm = 0.01m 2.900cm...
*July 1, 2013*

**Physics**

A uniform bar 1.000 meters in length and having a cross sectional area of 10.000cm^2 is subjected to a loading of 5750.000 Newtons. This load causes an extension to the bar length of 2.740mm. Calculate the stress and the strain produced. ANS 1 = kPa (Round to 3 decimal places...
*July 1, 2013*

**Physics**

The ultimate strength of a steel rod is 600.000MPa. if the factor of safety 3.000 is required, what is the maximum permissible load for the rod if it has a diameter of 6.000cm? ANS = kN (Round to 3 decimal place) Ultimate strength = 600.000MPa = 600000000Pa = 600 * 10^6 Factor...
*July 1, 2013*

**Physics**

1) Calculate the tensile stress in a 33.000mm diameter rod subjected to a pull of 30.000kN. ANS= MPa (Round to 3 decimal places) Diameter = 33.000mm = 0.033m 30.000kN = 30000N Area = pi*d^2/4 = 3.1416 * 0.033^2/ 4 = 3.1416 *0.001089/ 4 = 0.0034212024/ 4 = 8.553006*10^-4 = 0....
*July 1, 2013*

**Pyhsics(Stress&Strain)**

1) Calculate the tensile stress in a 33.000mm diameter rod subjected to a pull of 30.000kN. ANS= MPa (Round to 3 decimal places) Diameter = 33.000mm = 0.033m 30.000kN = 30000N Area = pi*d^2/4 = 3.1416 * 0.033^2/ 4 = 3.1416 *0.001089/ 4 = 0.0034212024/ 4 = 8.553006*10^-4 = 0....
*June 29, 2013*

**Physics(Sress&Strain)**

The ultimate strength of a steel rod is 600.000MPa. if the factor of safety 3.000 is required, what is the maximum permissible load for the rod if it has a diameter of 6.000cm? ANS = kN (Round to 3 decimal place) Ultimate strength = 600.000MPa = 600000000Pa = 600 * 10^6 Factor...
*June 29, 2013*

**Physics(Stress&Strain)**

A uniform bar 1.000 meters in length and having a cross sectional area of 10.000cm^2 is subjected to a loading of 5750.000 Newtons. This load causes an extension to the bar length of 2.740mm. Calculate the stress and the strain produced. ANS 1 = kPa (Round to 3 decimal places...
*June 29, 2013*

**Physics(Stress&Strain)**

A steel bar has a rectangular cross section measuring 1.000cm by 2.900cm and is 8.700m in length. A load of 189000.00N causes the bar to extend 1.500mm. Calculate the stress and strain produced. ANS 1 = kPa (3 decimal places) ANS 2 = (6 decimal places) 1.000cm = 0.01m 2.900cm...
*June 29, 2013*

**Physics(Stress&Strain)**

The Ultimate strength of a round steel rod is 630.000MPa. If a factor of safety of 5.000 is required, what is the maximum permissible load for the rod if it has a diameter of 3.000cm? ANS = kN (Round to 3 decimal places) Safety = 5.000 Rod diameter = 3.000cm = 0.03m Ultimate ...
*June 29, 2013*

**Physics (Stress & Strain)**

What is the factor of safety of a steel hanger having an ultimate strength of 630.000MPa and supporting a load of 65000.000N. The steel hanger in question has a cross sectional area of 5.300cm. ANS = (Round to 3 decimal places) Load = 65000.000N ultimate strength = 630.000MPa...
*June 29, 2013*

**Physics(Stress&Strain)**

What is the factor of safety of a steel hanger having an ultimate strength of 630.000MPa and supporting a load of 65000.000N. The steel hanger in question has a cross sectional area of 5.300cm. ANS = (Round to 3 decimal places) Load = 65000.000N ultimate strength = 630.000MPa...
*June 28, 2013*

**Physics (Stress&Strain)**

The Ultimate strength of a round steel rod is 630.000MPa. If a factor of safety of 5.000 is required, what is the maximum permissible load for the rod if it has a diameter of 3.000cm? ANS = kN (Round to 3 decimal places) Safety = 5.000 Rod diameter = 3.000cm = 0.03m Ultimate ...
*June 28, 2013*

**Physics (Stress & Strain)**

A steel bar has a rectangular cross section measuring 1.000cm by 2.900cm and is 8.700m in length. A load of 189000.00N causes the bar to extend 1.500mm. Calculate the stress and strain produced. ANS 1 = kPa (3 decimal places) ANS 2 = (6 decimal places) 1.000cm = 0.01m 2.900cm...
*June 28, 2013*

**Physics (Stress&Strain)**

A uniform bar 1.000 meters in length and having a cross sectional area of 10.000cm^2 is subjected to a loading of 5750.000 Newtons. This load causes an extension to the bar length of 2.740mm. Calculate the stress and the strain produced. ANS 1 = kPa (Round to 3 decimal places...
*June 28, 2013*

**Physics(Stress & Strain)**

1) Calculate the tensile stress in a 33.000mm diameter rod subjected to a pull of 30.000kN. ANS= MPa (Round to 3 decimal places) Diameter = 33.000mm = 0.033m 30.000kN = 30000N Area = pi*d^2/4 = 3.1416 * 0.033^2/ 4 = 3.1416 *0.001089/ 4 = 0.0034212024/ 4 = 8.553006*10^-4 = 0....
*June 28, 2013*

**Physics (Stress & Stain)**

Load = Allowable Strength * Area = Allowable Strength * (pi*d^2/4) =200*10^6*pi*.06^2/4 =200000000*3.1416*.0036/4 =2261952/4 =565488N =565.488kN Please check. Thank you.
*June 28, 2013*

**Physics (Stress & Stain)**

The ultimate strength of a steel rod is 600.000MPa. if the factor of safety 3.000 is required, what is the maximum permissible load for the rod if it has a diameter of 6.000cm? ANS = kN (Round to 3 decimal place) Ultimate strength = 600.000MPa = 600000000Pa = 600 * 10^6 Factor...
*June 28, 2013*

**Physics (Stress & Strain)**

The ultimate strength of a steel rod is 550.000MPa. If the factor of safety of 3.000 is required, what is the maximum permissible load for the rod if it has a diameter of 4.100cm? ANS = kN (Round to 3 decimal places) ultimate strength = 550.000MPa Factor of safety = 3.000 Rod ...
*June 28, 2013*

**Physics (Stress & Strain)**

What is the factor of safety of a steel hanger having an ultimate strength of 550.000MPa and supporting a load of 79000.000N. The steel hanger in question has a cross sectional area of 7.000cm^2. ANS = (Round to 3 decimal places) ultimate strength = 550.000MPa load = 79000....
*June 28, 2013*

**Physics (Stress & Strain)**

What is the factor of safety of a steel hanger having an ultimate strength of 550.000MPa and supporting a load of 79000.000N. The steel hanger in question has a cross sectional area of 7.000cm^2. ANS = (Round to 3 decimal places) ultimate strength = 550.000MPa load = 79000....
*June 27, 2013*

**Physics (Stress & Strain)**

The ultimate strength of a steel rod is 550.000MPa. If the factor of safety of 3.000 is required, what is the maximum permissible load for the rod if it has a diameter of 4.100cm? ANS = kN (Round to 3 decimal places) ultimate strength = 550.000MPa Factor of safety = 3.000 Rod ...
*June 27, 2013*

**Physics (Stress & Strain)**

A hole 3cm in diameter is to be punched out of a steel plate 8cm thick. The shear stress of the material is 670GPa. What load is required on the punch? 3cm = 0.03m 8cm = 0.08m Lateral surface area (As) = pi*D*h = 3.1416 * .03m * .08m = 3.1416 * .0024m = .007539m Force = Stress...
*June 26, 2013*

**Physics (Stress & Strain)**

A uniform bar 1.900 meters in length and having a cross sectional area of 11.000cm^2 is subject to a loading of 5450.000 Newtons. this load causes an extension to the bar length of 2.420mm. Calculate the stress and strain produced. 11cm = .11m 2.420mm = .02420m length = 1.9m ...
*June 26, 2013*

**Physics (Stress & Strain)**

A hole 3cm in diameter is to be punched out of a steel plate 8cm thick. The shear stress of the material is 670GPa. What load is required on the punch? 3cm = 0.03m 8cm = 0.08m Lateral surface area (As) = pi*D*h = 3.1416 * .03m * .08m = 3.1416 * .0024m = .007539m Force = Stress...
*June 26, 2013*

**Physics (Stress & Strain)**

A steel bar has a rectangular cross section measuring 1.000cm by 2.500cm and is 8.100m in length. A load of 150000.000N causes the bar to extend 2.000mm. Calculate the stress and strain. Area = l*w = 1.000cm*2.500cm = 2.500cm = 0.025m = 0.025m*8.100m = 0.2025m Stress = Load/ ...
*June 24, 2013*

**Physic (Stress & Strain)**

A uniform bar 1.900m in length and having a cross sectional area of 11.000cm^2 is subject to a loading of 5450.000N. this load causes an extension to the bar length of 2.420mm. Calculate the stress and the strain produced. Area = 11.000cm^2 = 11*10^-3m Load = 5450.000N Stress...
*June 24, 2013*

**Physics (Stress & Strain)**

Stress = force/ area = 87500N/ 5cm^2 = 17500 = 70000/ 17500 ANS = 4 Is that correct?
*June 24, 2013*

**Physics (Stress & Strain)**

What is the factor of safety for a steel hanger having an ultimate strength of 70000N per cm^2 and supporting a load of 87500N. The hanger has a cross sectional area of 5cm^2. Factor of safety = ultimate stress/ allowable stress = 70000/87500 = 0.8 Have I done this correctly?
*June 24, 2013*

**Physics (Stress & Strain)**

A hole 3cm in diameter is to be punched out of a steel plate 8cm thick. The shear stress of the material is 670GPa. What load is required on the punch? First find the area: A = pi*d^2/ 2 = 3.1416 * 3cm^2/ 2 = 3.1416 * 9/ 2 = 28.2744/ 2 = 7.0686cm Stress = 670GPa Second, covert...
*June 24, 2013*

**Math-algebra-Steve**

148
*June 24, 2013*

**Physics (Stress & Strain)**

1) Calculate the tension stress in a 21.00mm diameter rod subjected to a pull of 30.000kN. 21mm =0.021m 30.000kN = 30000N A = pi*d^2/4 = 3.1416 * (0.021)^2m^2/ 4 = 3.1416 * 0.000441/ 4 = 0.0013854456/ 4 = 0.0003463614 Stress = Load/ Area = 30000/ 0.0003463614 = 86614732.47N = ...
*June 23, 2013*

**Physics (Stress & Strain)**

What is the factory of safety of a steel hanger having an ultimate strength of 630.000MPa and supporting a load of 61000.000N. The steel hanger in question has a cross sectional area of 6.500cm^2. ANS = (Round to 3 decimal places) 6.500cm = 0.065m Load = 61000.000N Ultimate ...
*June 23, 2013*

**Physic (Stress & Strain)**

1) Calculate the tension stress in a 21.00mm diameter rod subjected to a pull of 30.000kN. 21mm =0.021m 30.000kN = 30000N A = pi*d^2/4 = 3.1416 * (0.021)^2m^2/ 4 = 3.1416 * 0.000441/ 4 = 0.0013854456/ 4 = 0.0003463614 Stress = Load/ Area = 30000/ 0.0003463614 = 86614732.47N = ...
*June 23, 2013*

**Physics (Stress & Strain)**

A uniform bar 1.900m in length and having a cross sectional area of 10.500cm^2 is subject to a loading of 5550.000N. This load causes an extension of the bar length of 1.100mm. Calculate the stress and strain produced. Stress = Load/ Area Strain = Change in length/ length ...
*June 23, 2013*

**Physics**

What is the factor of safety of a steel hanger having an ultimate strength of 550.000MPa and supporting a load of 64000.000N. The steel hanger in question has a cross sectional area of 7.600cm^2 ANS= Round to 3 decimal places Safety factor = design load/ load = 550.000MPa = ...
*June 22, 2013*

**Physics**

What is the factor of safety of a steel hanger having an ultimate strength of 550.000MPa and supporting a load of 64000.000N. The steel hanger in question has a cross sectional area of 7.600cm^2 ANS= Round to 3 decimal places Safety factor = design load/ load = 550.000MPa = ...
*June 22, 2013*

**Physics**

A tie bar is made of a material having a tensile strength of 231MPa and has to carry a load of 225kN. What is the diameter of the bar if a factor of safety of 7 is applied? Where do I begin?
*June 22, 2013*

**Physics**

Safety factor = design load/ load = 550.000MPa = 550000000Pa = 550000000/ 64000 = 8593.75 tension design = 550MPa * 0.076M^2 = 18.000 Have I messed thing up?
*June 21, 2013*

**Physics**

Sorry i missed this part of the question: The steel hanger in question has a cross sectional area of 7.600cm^2
*June 21, 2013*

**Physics**

What is the factor of safety of a steel hanger having an ultimate strength of 550.000MPa and supporting a load of 64000.000N. ANS= Round to 3 decimal places Where do I begin?
*June 21, 2013*

**Physics**

A tie bar is made of a material having a tensile strength of 231MPa and has to carry a load of 225kN. What is the diameter of the bar if a factor of safety of 7 is applied? Where do I begin?
*June 21, 2013*

**math**

634
*June 21, 2013*

**Physics**

What is the factor of safety of a steel hanger having an ultimate strength of 550.000MPa and supporting a load of 64000.000N. The steel hanger in question has a cross sectional area of 7.600cm^2. Answer = (Round to 3 decimal places)
*June 8, 2013*

**Physics**

1. A train traveling at a speed of 110km/h decelerates for 20 seconds. During that time it travels 450 meters. A) What is the acceleration? = m/s^2 B) What is its speed after deceleration? = km/h Initial velocity (u) = 110km/h Time (t) = 20s Distance = 450m Using this formula ...
*June 7, 2013*

**Physics**

1. A train traveling at a speed of 110km/h decelerates for 20 seconds. During that time it travels 450 meters. A) What is the acceleration? = m/s^2 B) What is its speed after deceleration? = km/h I need help with which formula to use and how to use it to find the correct ...
*June 6, 2013*

**Physics**

1. A train traveling at a speed of 150km/h decelerates for 15 seconds. During that time it travels 400 meters. A) What is the acceleration? = m/s^2 B) What is its speed after deceleration? = km/h I need help with which formula to use and how to use it to find the correct ...
*June 6, 2013*

**Physics**

1. A train traveling at a speed of 110km/h decelerates for 20 seconds. During that time it travels 450 meters. A) What is the acceleration? = m/s^2 B) What is its speed after deceleration? = km/h I need help with which formula to use and how to use it to find the correct ...
*June 6, 2013*

**Physics**

1. A train traveling at a speed of 150km/h decelerates for 15 seconds. During that time it travels 400 meters. A) What is the acceleration? = m/s^2 B) What is its speed after deceleration? = km/h I need help with which formula to use and how to use it to find the correct ...
*June 6, 2013*

**Physics**

1. A train traveling at a speed of 150km/h decelerates for 15 seconds. During that time it travels 400 meters. A) What is the acceleration? = m/s^2 B) What is its speed after deceleration? = km/h I need help with which formula to use and how to use it to find the correct ...
*June 5, 2013*

**Physics**

1. A car is traveling at 85km/h, find the following: A) Its speed in m/s u = 85km/h = 85km/h * 1000m/km/ 60s/min * 60min/m/h = 85000/ 3600 = 23.61 ANS = 23.61m/s B) How many meters will it travel in 8 seconds = 23.61m/s * 8s ANS = 188.88m C) How long (in seconds) will it take ...
*June 5, 2013*

**Velocity**

An object is thrown directly downwards from a height of 60m with an initial velocity of 11m/s. What will be its velocity on impact? v^2 = U^2 + 2as a= 9.81m/s^2 u = 11m/s s = 60m v^2 = 11m/s + 2 * 9.81m/s^2 * 60 = 1188.2m^2/s^2 = square root of 1188.2m^2/s^2 = 34.47m/s Is this...
*June 3, 2013*

**Acceleration**

Which of the following statements about acceleration is true. Acceleration is: A) time/ velocity B) force/ time C) the rate of change of displacement D) time/ force E) the result of an unbalanced force exerted on a body. ANS = E Is this correct??
*June 3, 2013*

**Acceleration**

Which of the following statements about acceleration is true. Acceleration is: A) time/ velocity B) force/ time C) the rate of change of displacement D) time/ force E) the result of an unbalanced force exerted on a body. ANS = E Is this correct??
*June 3, 2013*

**Linear velocity/ acceleration**

A body moves 180 degrees from point A to point B in a semi circle. The diameter of the semi circle is 2m. Calculate the distance traveled between point A and point B. C = pi * d = 3.1416 * 2m = 6.283 = 6.283/ 2 = 3.141 ANS = 3.141 Is this done correctly??
*June 3, 2013*

**Physics**

Camera lenses are described in terms of their focal length. A 50.0 mm lens has a focal length of 50.0 mm (a) A camera with a 50 mm lens is focused on an object 4.0 m away. Locate the image. (b) A 980 mm lens is focused on an object 125 m away. Locate the image.
*May 21, 2013*

**Physics**

A wheel and axle arrangement has an axle with a diameter of 25cm. A rope is wrapped around the axle and is supporting a load of 2 tonnes. 1)If the mechanical advantage of the system is 30, determine the diameter of the wheel. ANSWER = m 2)Determine the effort required to ...
*May 20, 2013*

**Physics- Simple machines**

A wheel and axle arrangement has an axle with a diameter of 25cm. A rope is wrapped around the axle and is supporting a load of 2 tonnes. 1) If the mechanical advantage of the system is 30, determine the diameter of the wheel___________m. 2)Determine the effort required to ...
*May 12, 2013*

**Physics**

Simple Machines A simple wheel and axle has a wheel diameter of 2.3m and an axle diameter of 92mm. If an effort of 26N is required to raise a mass of 16 kg what is the efficiency of the machine? A- 2.46% B- 95.3% C- 24.1% D- 20.1% E- 65.5% I am having a problem determining ...
*May 11, 2013*

**MATH**

the area of a certain rectangle is 288yd squared. the perimeter is 68yd. if you double the length and width,what will be the area and perimeter of the new rectangle?
*May 9, 2013*

**Physics**

300N 200N ? |- 8m -|- 5m - |_____________________________________| R2 10M | 10m R1 A beam having a length of 20 metres is pivoted at its mid point. A 200 newton load is located at a point 5 metres from the right hand end of the beam. A 300 newton load is located at a point 8 ...
*May 4, 2013*

**Physics**

A beam having a length of 20 metres is pivoted at its mid point. A 200 newton load is located at a point 5 metres from the right hand end of the beam. A 300 newton load is located at a point 8 metres from the right hand end. In order to be in equilibrium, what load is required...
*May 4, 2013*

**physics**

A body of 150kg is located at a height of 45m. If allowed to fall freely to the ground. What will be its velocity on impact?
*April 29, 2013*

**Physics**

A body is moving north at 12m/s and 5 seconds later it has a velocity of 30m/s in the same direction. What was the acceleration of the body?
*April 27, 2013*

**Physics**

A Body with a mass of 150kg is located at a height of 45 m. If allowed to fell freely to the ground. What will be its velocity on impact? KE= 1/2*m*v^2 PE= m*g*h 1/2*m*v^2= m*g*h 1/2*150kg*45m*v^2= 150kg*9.81*45m 3375* v^2= 66217.5 3375*v^2/3375= 66217.5/3375 v^2= 19.62m/s Is ...
*April 27, 2013*