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miskamm [114]
3 years ago
10

A COPPER WIRE OF LENGTH 4M AND AREA OF CROSS-SECTION 1.2CM^2 IS STRECHED WITH A FORCE 4.8×10^3 N . IF YOUNGS MODULAS FOR COPPER

IS 1.2×10^11 N/M WHAT IS THE INCREMENT OF LENGTH OF THE WIRE ?????..
NEED THE ANSWER SOON MAYBE IN SOME MINUTES



HOPE YOU WILL ANSWER ​
Physics
1 answer:
VARVARA [1.3K]3 years ago
6 0

Stress = force / area

= 4.8 x 10³ / 1.2 x 10^-4

= 4 x 10⁷ N /m²

YOUNGS MODULAS = stress / strain

= 4 x 10⁷ / 1.2 x 10^11

= 3.3 x 10^-4

INCREMENT OF LENGTH = longitudinal length x intitial length

= ( 3.3 x 10^-4 ) x 4

= 13.2 x 10^-4 m

= 13.2 mm

mark ‼️ it brainliest if it helps you ❤️

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The magnitude of displacements a and b are 3m and 4m, respectively, c=a+b. What is the magnitude of c if the angel between a and
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Explanation:

Given:

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Therefore, the magnitude of vector 'c' is 7 m when angle between 'a' and 'b' is 0°.

(b)

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So, magnitude of 'c' when 'a' and 'b' are in opposite direction is:

|\overrightarrow c|=||\overrightarrow a|-|\overrightarrow b||\\\\|\overrightarrow c|=|3 - 4| = 1\ m

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