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Using this equation for an average speed of 80 kilometres per hour on a 4-hour trip, the distance covered is found to be 320 kilometres. Expressed in graphical language, the slope of a tangent line at any point of a distance-time graph is the instantaneous speed at this point, while the slope of a chord line of the same graph is the average ...
The first equation shows that, after one second, an object will have fallen a distance of 1/2 × 9.8 × 1 2 = 4.9 m. After two seconds it will have fallen 1/2 × 9.8 × 2 2 = 19.6 m; and so on. On the other hand, the penultimate equation becomes grossly inaccurate at great distances.
When the second hand reaches the point on the scale where the speed indicated equals the speed of the vehicle, one unit of distance (miles if speed is miles per hour, kilometres if kilometres per hour, etc.) has been covered. For example, if you travel at a constant 80 mph (or at 80 km/h), then the distance travelled while the second hand ...
Illustration of the walking example. In elementary algebra, the unitary method is a problem-solving technique taught to students as a method for solving word problems involving proportionality and units of measurement. It consists of first finding the value or proportional amount of a single unit, from the information given in the problem, and ...
Knots tied at a distance of 47 feet 3 inches (14.4018 m) from each other, passed through a sailor's fingers, while another sailor used a 30-second sand-glass (28-second sand-glass is the currently accepted timing) to time the operation. [9] The knot count would be reported and used in the sailing master's dead reckoning and navigation.
Once this and the conversion factor for seconds per hour have been multiplied by the original fraction to cancel out the units mile and hour, 10 miles per hour converts to 4.4704 metres per second. As a more complex example, the concentration of nitrogen oxides (NO x) in the flue gas from an industrial furnace can be converted to a mass flow ...
Escape speed at a distance d from the center of a spherically symmetric primary body (such as a star or a planet) with mass M is given by the formula [2] [3] = = where: G is the universal gravitational constant (G ≈ 6.67 × 10 −11 m 3 ⋅kg −1 ⋅s −2 [4])
That is, 7.92 units of distance are equivalent to 1 unit of climb. For convenience an 8 to 1 rule can be used. So, for example, if a route is 20 kilometres (12 mi) with 1600 metres of climb (as is the case on leg 1 of the Bob Graham Round, Keswick to Threlkeld), the equivalent flat distance of this route is 20+(1.6×8)=32.8 kilometres (20.4 mi ...