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Flura [38]
3 years ago
6

The following arbitrary measurements are made and the errors sited are the aximum errors A = 15.21 +0.01, B = 10.82 +0.05, C = 1

1.00+ 0.03. If D= A + B + C; (a) Calculate the maximum error in D. (b) if the errors sited are standard errors, calculate the standard error in D.​
Physics
1 answer:
Molodets [167]3 years ago
3 0

Maximum error in the result of the sum of measurement is equal to the sum absolute error of the individual observed measurements

(a) The maximum error in D is 0.09

(b) The standard error in D is approximately 0.034

The procedure for arriving at the above values is as follows;

The given measurements and the sited errors are;

A = 15.21 + 0.01

B = 10.82 + 0.05

C = 11.00 + 0.03

D = A + B + C

(a) Required parameter;

To calculate the maximum error in D

The equation for the propagation of error in addition is presented as follows;

Given that we have;

x = a + b

Therefore;

x + ±Δx = (a ± Δa) + (b ± Δb) = a + b ± (Δa + Δb)

∴ Δx = Δa + Δb

From the above formula, we have;

Where;

D = A + B + C

The maximum error in D = The sum of the maximum error in A, B, C

∴ The maximum error in D = 0.01 + 0.05 + 0.03 = 0.09

(b) Required parameter:

To find the standard error in D

The standard error is the sampling distribution's standard deviation, SD

Variance = SD²

The combined variance, SD² = The sum of the squares of individual standard deviations

Given that the standard errors represents the standard deviation, we get;

The combined variance, SD² = 0.01² + 0.05² + 0.03²

The combined variance, SD = √(0.01² + 0.05² + 0.03²) = 0.059

Standard \ error = \dfrac{SD}{\sqrt{n} }

Where n = 3, for the three measurement, we get;

Standard \ error = \dfrac{\sqrt{0.01^2 + 0.05^2 + 0.03^2} }{\sqrt{3} }  \approx 0.034

The standard error in D is approximately 0.034

Learn more about maximum error and standard error here:

brainly.com/question/13106593

brainly.com/question/17164235

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Answer:

A

Explanation:

In Column A, the pattern is 1², 2², 3², etc.  So x(t) = t², which is quadratic.

In Column B, the pattern is 2*1, 2*2, 2*3, etc.  So x(t) = 2t, which is linear.

In Column C, the pattern is 9*1, 9*2, 9*3, etc.  So x(t) = 9t, which is linear.

In Column D, the pattern is 1/1, 1/2, 1/3, etc.  So x(t) = 1/t, which is reciprocal.

In Column E, the pattern is 1/1², 1/2², 1/3², etc.  So x(t) = 1/t², which is reciprocal.

5 0
4 years ago
A 20-kg boy slides down a smooth, snow-covered hill on a plastic disk. The hill is at a 10° angle to the horizontal, and the slo
zvonat [6]

Answer:

13 m/s

Explanation:

I assume we are ignoring friction.

The boy's PE will all be converted to KE at the bottom of the hill.

to find PE = mgh   we need to know h

   h = 50 sin 10 = 8.68 meters

     then:    PE = 20 * 9.81 * 8.68 =<u> 1703.49</u> j

KE = 1/2 m v^2 = <u>1703 .49</u>

            v = 13 m/s

7 0
2 years ago
At takeoff, an aircraft travels at 62 m/s, so that the air speed relative to the bottom of the wing is 62 m/s. Given the sea lev
olganol [36]

Answer:

the aircraft must travel at a speed of <em>73.4 m/s</em> in order to create the ideal lift.

Explanation:

We will use Bernoulli's theorem in order to determine the pressure lift:

ΔP = 1/2 (ρ)(v₂² - v₁²)

the generated pressure lift is ΔP = 1000 N/m²

Therefore,

1000 = 1/2(ρ)(v₂² - v₁²)

v₂² - v₁² = 2000 / ρ

v₂² = (2000 N/m² / 1.29 kg/m³) + (62 m/s)²

v₂ = √[ (2000 N/m² / 1.29 kg/m³) + (62 m/s)² ]

<em>v₂ = 73.4 m/s </em>

<em></em>

Therefore, the aircraft must travel at a speed of <em>73.4 m/s</em> in order to create the ideal lift.

5 0
4 years ago
A meteoroid is in a circular orbit 600 km above the surface of a distant planet. The planet has the same mass as Earth but has a
AVprozaik [17]
<h2>Answer:</h2><h2>The acceleration of the meteoroid due to the gravitational force exerted by the planet = 12.12 m/s^{2}</h2>

Explanation:

A meteoroid is in a circular orbit 600 km above the surface of a distant planet.

Mass of the planet = mass of earth = 5.972 x 10^{24} Kg

Radius of the earth = 90% of earth radius = 90% 6370 = 5733 km

The acceleration of the meteoroid due to the gravitational force exerted by the planet = ?

By formula, g = \frac{GM}{r^{2} }

where g is the acceleration due to the gravity

G is the universal gravitational constant = 6.67 x 10^{-11} m^{3} kg^{-1} s^{-2}

M is the mass of the planet

r is the radius of the planet

Substituting the values, we get

g =  \frac{(6.67 * 10^{-11}) (5.972 * 10^{24})  }{5733^{2} }

g = 12.12 m/s^{2}

The acceleration of the meteoroid due to the gravitational force exerted by the planet = 12.12 m/s^{2}

6 0
4 years ago
Neon atoms at 245 K pass through a fan that gives each mole of neon gas an additional kinetic energy of 16.0 J. Part A What is t
hjlf

Answer:

246.28 K

Explanation:

The total energy of one mole of gas molecules can be calculated by the formula given below

E = \frac{3}{2}\times R\times T

Where R is gas constant and T is absolute temperature.

Put the value of R as 8.314 and temperature as 245 , we get

E = \frac{3}{2}\times 8.314\times 245

= 3055.4 J

Add 16 j to it

Total energy of gas molecules = 3055.4 + 16 = 3071.4 J.

If T be the temperature after addition of energy then

\frac{3}{2}\times 8.314\times T = 3071.4

T =\frac{2\times 3071.4}{3\times 8.314}

T = 246.28 K

7 0
3 years ago
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