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solong [7]
3 years ago
13

Using Hooke's law F spring = k dela x, find elastic constant of a spring that stretches 2cm when a 4 newton force is applied to

it
Answers in the picture

Physics
1 answer:
photoshop1234 [79]3 years ago
6 0

Answer: <em>2 N/cm  (B)</em>

Given that:

Spring stretches = 2 cm ,

Force  = 4 N,

<em>determine </em>

            the spring constant/elastic constant ( k)= ?

            We know that from hook's law, <em>F = k.x</em>  N

                                                                 4 = k × 2

                                                                => k  =4/2

                                                                         = 2 N/cm

The elastic constant is 2 N/cm

           

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Answer:

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Explanation:

It is given that,

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Time, t = 1.1 min = 66 seconds

Let W is the work done by the tension. It is equal to the product of force and displacement. It is given by :

W=F\times d

Since, d=vt

W=F\times v\times t

W=465\ N\times 4.6\ m/s\times 66\ s

W = 141174 Joules

So, the work is done by the tension is 141174 Joules. Hence, this is the required solution.

6 0
3 years ago
Consider a product with three components in​ series, with reliabilities of​ 0.90, 0.80, and 0.99 for components​ A, B, and​ C, r
Feliz [49]

Answer:0.853

Explanation:

Given

Reliability of A is 0.90

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i.e. B has a component in parallel to it.

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B actual reliability is

R_b=1-\left ( 1-0.8\right )\left ( 1-0.8\right )

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Thus R_{net} is given by

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Answer:

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We can calculate the parallel one since it's the one that affects the force required to push up

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Since customer would not complain if the force is no more than 20N

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