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3 years ago

A spring of original length 3.0 cm is extended to a total length of 5.0 cm by a force of 8.0N

Physics

2 answers:

miss Akunina [59]3 years ago

8 0

Answer:

12.0N

Explanation:

This problem is solved using Hooke's law which states that the extension or compression experienced by an elastic material is proportional the the force or load applied provided that the elastic limit is not exceeded.

$F=ke..................(1)$

where F is the force applied, k is the elastic constant and e is the extension of the material.

$e=l_1-l_o..................(2)$

were $l_o$ is the original length of the material and $l_1$ is its total length after extension.

Given;

F = 8.0N

$l_o=3cm\\l_1=5cm\\k=?$

therefore,

$e=5-3\\e=2cm$

substituting all necessary values into equation (1), we get;

$8=k*2\\k=\frac{8}{2}\\k=4N/cm$

The force required to give the spring a total length of 6cm is thus calculated;

in this case,

$l_1=6cm\\therefore\\e=6-3\\e=3cm$

hence F is given as follows;

$F=(4N/cm) *3cm\\F=12.0N$

hjlf3 years ago

6 0

4848484 5 5 5 5 5 =++++

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3 years ago

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3 years ago

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2 years ago

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GenaCL600 [577]

Answer:

length of the ladder is 13.47 feet

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angle between ladder and the wall is 26.95°

Explanation:

given data

height h = 12 feet

angle 63°

to find out

length of the ladder ( L) and length of wall to ladder ( A) and angle between ladder and the wall

solution

we consider here angle between base of wall and floor is right angle

we apply here trigonometry rule that is

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and

we can say

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and

we say here

tanθ = 6.10 / 12

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so angle between ladder and the wall is 26.95°

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3 years ago