Which best describes the motion of the object between 1

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

Molodets [167]

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:

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Consider that a clay model of a tiger has a mass of 0.195 kg and glides on ice at a speed of 0.75 m/s. It hits another clay mode

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Answer:

The final velocity after the collision is 0.27 m/s.

Explanation:

Given that,

Mass of tiger, m = 0.195 kg

Initial speed of tiger model, v = 0.75 m/s

Mass of another clay model, m' = 0.335 kg

Initially, second model is at rest, v' = 0

We need to find the final velocity after the collision. It is a case of inelastic collision. Using the conservation of linear momentum as :

$mv+m'v'=(m+m')V\\\\V=\dfrac{mv+m'v'}{(m+m')}\\\\V=\dfrac{0.195\times 0.75}{(0.195 +0.335 )}\\\\V=0.27\ m/s$

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