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OLga [1]
3 years ago
12

​Hooke's Law. The distance d when a spring is stretched by a hanging object varies directly as the weight w of the object. If th

e distance is 10 cm when the weight is 3 ​kg, what is the distance when the weight is 8 ​kg?
Physics
1 answer:
nadezda [96]3 years ago
7 0

Answer:

distance when the weight is 8 ​kg is 26.66 cm

Explanation:

given data

distance d2 = 10 cm

weight w2 = 3 ​kg

weight w1 = 8 kg

to find out

distance when the weight is 8 ​kg

solution

we consider here distance d1 when weight is 8 kg

so equation will be

d1/d2 = w1/w2

d/10 = 8/ 3

so d = 8/3 × 10

so d = 26.66

distance when the weight is 8 ​kg is 26.66 cm

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Answer:

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Explanation:

Given;

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original length of invar, L₁ = 1.00 m

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Where;

ΔL is difference in length

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ΔL =  L₁αθ

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L₁ = 1.00 m

θ = 20.5°C

\alpha = \frac{\gamma}{3} = \frac{2.7*10^{-6}}{3} = 0.9*10^{-6}/^oC

ΔL  = 1 x 0.9 x 10⁻⁶ x 20.5 = 1.845 x 10⁻⁵ m

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3 years ago
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