HP 35s Handbuch

hp calculators

HP 35s General applications - Part 3

Other applications

Practice solving problems

- Application 1: True Heading

- Application 2: Thrown object

- Application 3: Gas Pressure

hp calculators

HP 35s General applications – part 3

hp calculators - 2 - HP 35s General applications – part 3- Version 1.0

General applications

This training aid will illustrate the application of the HP 35s calculator to several problems in other areas. These

examples are far from exhaustive, but do indicate the incredible flexibility of the HP 35s calculator.

Practice solving problems

Application 1: True Heading

Example 1: Before a plane takes off, a preflight plan must be filed indicating the proposed trip. In these plans, the

known data are true course (TC), wind direction (WD), wind velocity (WV) and true air speed (TAS). Find

the true heading of a planned flight, if TC = 80 degrees, WV = 55 mph, TAS = 180 mph, and the WD is

toward 95 degrees.

The true heading = TC – ASIN ( WV x SIN ( WD – TC ) / TAS )

Solution: Place the 35s into degrees mode by pressing 91

In RPN mode: 80Ï55Ï95Ï80Ãk180¯

¸ºiÃ

In algebraic mode: 80Ãºi55¸k95Ã80Õ¯180

Figure 1

Answer: 75.46 degrees.

Application 2: Thrown object

Example 1: If a ball is thrown straight upward with a velocity of 95 feet per second, what is the velocity and height of the

object after 2 seconds? Use a value of 32.2 feetper second persecond for g.

Velocity = OriginalVelocity – g x time

Height = Original Velocity x time – ½ x g x t2

Solution: In RPN mode: 95Ï32Ë2Ï2¸Ã (Velocity at t = 2 seconds)

95Ï2¸0Ë5Ï

32Ë2¸2º=Ã (Height at t = 2 seconds)

In algebraic mode: 95Ã32Ë2¸2Ï (Velocity at t = 2 seconds)

95¸2Ã0Ë5¸

32Ë2¸º=2Ï (Height at t = 2 seconds)

hp calculators

HP 35s General applications – part 3

hp calculators - 3 - HP 35s General applications – part 3- Version 1.0

Figure 2

Figure 3

Answer: After 2 seconds, the object is at a height of 125.6 feet and traveling at a velocity of 30.6 feet per second.

Application 3: Gas pressure

Example 1: The internalpressure of a tank of gas is 1100 psi at room temperature (298 degrees Kelvin). What is the

internal pressure if the temperature rises by 35 degrees Celsius?

New Pressure = Old Pressure x New Temperature ÷ Old Temperature

Solution: In RPN mode: 1100Ï298Ï35Ù¸298

In algebraic mode: 1100¸4298Ù35Õ¯298Ï

Figure 4

Answer: The new pressure is 1229 psi.

Andere Handbücher für 35s

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