# November 2018

## What is the Coefficient of the 8th Term of the Expansion of (2x - 1/x)^10?

**Problem**

In the expansion of (2*x* - 1/*x*)^{10}, find the coefficient of the 8^{th} term.

A. 980 | C. 960 |

B. 970 | D. 990 |

## How Far An Object Has Fallen If Its Velocity Is 80 Feet Per Second

**Problem**

The formula $v = \sqrt{2gh}$ give the velocity, in feet per second, of an object when it falls *h* feet accelerated by gravity *g*, in feet per second squared. If *g* is approximately 32 feet per second squared, find how far an object has fallen if its velocity is 80 feet per second.

A. 80 feet | C. 70 feet |

B. 100 feet | D. 90 feet |

## How Long Would it Take to Fly From Earth to Jupiter?

**Problem**

Earth is approximately 93,000,000.00 miles from the sun, and the Jupiter is approximately 484,000,900.00 miles from the sun. How long would it take a spaceship traveling at 7,500.00 mph to fly from Earth to Jupiter?

A. 9.0 years | C. 6.0 years |

B. 5.0 years | D. 3.0 years |

## Volume of Inflating Spherical Balloon as a Function of Time

**Problem**

A meteorologist is inflating a spherical balloon with a helium gas. If the radius of a balloon is changing at a rate of 1.5 cm/sec., express the volume *V* of the balloon as a function of time *t* (in seconds). Hint: Use composite function relationship *V*_{sphere} = 4/3 π*r*^{3} as a function of *x* (radius), and *x* (radius) as a function of *t* (time).

A. V(t) = 5/2 πt^{3} |
C. V(t) = 9/2 πt^{3} |

B. V(t) = 7/2 πt^{3} |
D. V(t) = 3/2 πt^{3} |

## Smallest Triangular Portion From A Square Lot

**Problem**

A farmer owned a square field measuring exactly 2261 m on each side. 1898 m from one corner and 1009 m from an adjacent corner stands Narra tree. A neighbor offered to purchase a triangular portion of the field stipulating that a fence should be erected in a straight line from one side of the field to an adjacent side so that the Narra tree was part of the fence. The farmer accepted the offer but made sure that the triangular portion was a minimum area. What was the area of the field the neighbor received and how long was the fence? Hint: Use the Cosine Law.

A. A = 972,325 m^{2} and L = 2,236 m |

B. A = 950,160 m^{2} and L = 2,122 m |

C. A = 946,350 m^{2} and L = 2,495 m |

D. A = 939,120 m^{2} and L = 2,018 m |

## Longest Day of the Year: Summer Solstice

**Problem**

The number of hours daylight, *D*(*t*) at a particular time of the year can be approximated by

for *t* days and *t* = 0 corresponding to January 1. The constant *K* determines the total variation in day length and depends on the latitude of the locale. When is the day length the longest, assuming that it is NOT a leap year?

A. December 20 | C. June 20 |

B. June 19 | D. December 19 |

## Velocity of Separation: How fast is the distance between two cars changing?

**Problem**

A Toyota Land Cruiser drives east from point *A* at 30 kph. Another car, Ford Expedition, starting from *B* at the same time, drives S30°W toward *A* at 60 kph. *B* is 30 km from *A*. How fast in kph is the distance between two cars changing after 30 minutes? Hint: Use the Cosine Law.

A. 70 kph | C. 55 kph |

B. 80 kph | D. 60 kph |

## Simply Supported Beam with Support Added at Midspan to Prevent Excessive Deflection

**Situation**

A simply supported beam has a span of 12 m. The beam carries a total uniformly distributed load of 21.5 kN/m.**1.** To prevent excessive deflection, a support is added at midspan. Calculate the resulting moment (kN·m) at the added support.

A. 64.5 | C. 258.0 |

B. 96.8 | D. 86.0 |

**2.** Calculate the resulting maximum positive moment (kN·m) when a support is added at midspan.

A. 96.75 | C. 108.84 |

B. 54.42 | D. 77.40 |

**3.** Calculate the reaction (kN) at the added support.

A. 48.38 | C. 161.2 |

B. 96.75 | D. 80.62 |

## Limit the Deflection of Cantilever Beam by Applying Force at the Free End

**Situation**

A cantilever beam, 3.5 m long, carries a concentrated load, *P*, at mid-length.

**Given:**

*P*= 200 kN

Beam Modulus of Elasticity,

*E*= 200 GPa

Beam Moment of Inertia,

*I*= 60.8 × 10

^{6}mm

^{4}

**1.** How much is the deflection (mm) at mid-length?

A. 1.84 | C. 23.50 |

B. 29.40 | D. 14.70 |

**2.** What force (kN) should be applied at the free end to prevent deflection?

A. 7.8 | C. 62.5 |

B. 41.7 | D. 100.0 |

**3.** To limit the deflection at mid-length to 9.5 mm, how much force (kN) should be applied at the free end?

A. 54.1 | C. 129.3 |

B. 76.8 | D. 64.7 |

## Support Added at the Midspan of Simple Beam to Prevent Excessive Deflection

**Situation**

A simply supported steel beam spans 9 m. It carries a uniformly distributed load of 10 kN/m, beam weight already included.

**Given Beam Properties:**

Area = 8,530 mm

^{2}

Depth = 306 mm

Flange Width = 204 mm

Flange Thickness = 14.6 mm

Moment of Inertia,

*I*= 145 × 10

_{x}^{6}mm

^{4}

Modulus of Elasticity,

*E*= 200 GPa

1. What is the maximum flexural stress (MPa) in the beam?

A. 107 | C. 142 |

B. 54 | D. 71 |

2. To prevent excessive deflection, the beam is propped at midspan using a pipe column. Find the resulting axial stress (MPa) in the column

**Given Column Properties:**

Outside Diameter = 200 mm

Thickness = 10 mm

Height = 4 m

A. 4.7 | C. 18.8 |

B. 9.4 | D. 2.8 |

3. How much is the maximum bending stress (MPa) in the propped beam?

A. 26.7 | C. 15.0 |

B. 17.8 | D. 35.6 |