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3 years ago

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Calculate the value of the error (only) with one decimal place for: (21+1-14%) x (14 +1-9%) Please enter the answer without +- s

ign

1 answer:

Brrunno [24]3 years ago

8 0

Answer:

6.9.

Explanation:

In case of product of two physical quantities with errors , the rule is that the product will contain error equal to the sum of percentage errors of the two physical quantities. Foe example P contains 2% error and Q contains 3% error then P X Q will contain 2 +3 = 5 % error.

(21 ± 1.14 %) x (14±1.9%)

Total percentage error in the result of product

1.14 + 1.19 = 2.33 %.

Produce value = 21 x 14 = 294

percent of error in it = 2.33%

value of error

= 2.33% of 294

= 6.8502 or 6.9.

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Answer:

$\frac{1}{10}$ M

Explanation:

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As the <u>asteroid is travelling directly towards the center of the Earth</u>, after impact ,it <u>does not impose any torque on earth's rotation,</u> So angular momentum of earth is conserved

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$W_{1}$ is the angular velocity of earth about an axis passing through the center of earth before impact
$I_{2}$ is moment of interia of earth and asteroid system
$W_{2}$ is the angular velocity of earth and asteroid system about the same axis

let $W_{1}=W$

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therefore $W_{1}$ $=$ $\frac{4}{5} \times W_{1}$

$I_{1}=\frac{2}{5} \times M\times R^{2}$ (moment of inertia of solid sphere)

where M is mass of earth

R is radius of earth

$I_{2}=\frac{2}{5} \times M\times R^{2}+M_{1}\times R^{2}$

(As given asteroid is very small compared to earth, we assume it be a particle compared to earth, therefore by parallel axis theorem we find its moment of inertia with respect to axis)

where $M_{1}$ is mass of asteroid

⇒ $\frac{2}{5} \times M\times R^{2} \times W_{1}=}(\frac{2}{5} \times M\times R^{2}$ $+$ $M_{1}\times R^{2})\times(\frac{4}{5} \times W_{1})$

$\frac{1}{2} \times M\times R^{2}$ = $(\frac{2}{5} \times M\times R^{2}$ + $M_{1}\times R^{2})$

$M_{1}\times R^{2}=$ $\frac{1}{10} \times M\times R^{2}$

⇒ $M_{1}=}$ $\frac{1}{10} \times M$

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Answer:

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Answer:

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