Answer:
Systematic error can be corrected using calibration of the measurement instrument, while random error can be corrected using an average measurement from a set of measurements.
Explanation:
Random errors lead to fluctuations around the true value as a result of difficulty taking measurements, whereas systematic errors lead to predictable and consistent departures from the true value due to problems with the calibration of your equipment.
Systematic error can be corrected, by calibration of the measurement instrument. Calibration is simply a procedure where the result of measurement recorded by an instrument is compared with the measurement result of a standard value.
Random error can be corrected using an average measurement from a set of measurements or by Increasing sample size.
There are many factors that determine if an aircraft can operate from a given airport. Of course the availability of certain services, such as fuel, access to air stairs and maintenance are all necessary. But before considering anything else, one must determine if the plane can physically land at an airport, and equally as important, take off.
What is the minimum runway length that will serve?
Looking at aerial views of runways can lead some to the assumption that they are all uniform, big and appropriate for any plane to land. This couldn’t be further from the truth.
A given aircraft type has its own individual set of requirements in regards to these dimensions. The classic 150’ wide runway that can handle a wide-body plane for a large group charter flight isn’t a guarantee at every airport. Knowing the width of available runways is important for a variety of reasons including runway illusion and crosswind condition.
Runways also have different approach categories based on width, and have universal threshold markings that indicate the actual width.
To learn more about runway
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A boy throws a ball and accidentally breaks a window. The momentum of the ball and all the pieces of glass taken together after the collision is the same as <span>the momentum of the ball before the collision. I think you forgot to give the choices along with the question. I hope that the answer has come to your great help.</span>
Volume = mass / density = 45.6/10.5 = .... L
Let M = mass of the skier,
v2 = his speed at the end of the track.
By conservation of energy,
1/2 Mv^2 = 1/2 Mv2^2 + Mgh
Dividing by M,
1/2 v^2 = 1/2 v2^2 + gh
Multiplying by 2,
v^2 = v2^2 + 2gh
Or v2^2 = v^2 - 2gh
Or v2^2 = 4.8^2 - 2 * 9.8 * 0.46
Or v2^2 = 23.04 - 9.016
Or v2^2 = 14.024 m^2/s^2-----------------------------(1)
In projectile motion, launch speed = v2
and launch angle theta = 48 deg
Maximum height
H = v2^2 sin^2(theta)/(2g)
Substituting theta = 48 deg and value of v2^2 from (1),
H = 14.024 * sin^2(48 deg)/(2 * 9.8)
Or H = 14.024 * 0.7431^2/19.6
Or H = 14.024 * 0.5523/19.6
Or H = 0.395 m = 0.4 m after rounding off
Ans: 0.4 m
The answer in this question is 0.4 m
