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ankoles [38]
3 years ago
15

Suppose a spring has a relaxed length of 28.3 cm. The simulation refers to this as the natural length. This is the length of the

spring when it is not stretched nor compressed by any force. When you hang a mass of 550 g on the vertical spring, the spring stretches to a new length of 38.2 cm. What is an estimate of the spring constant, k, of the spring? (Enter your answer in N/m.) Measuring only one value is not a good way to get the value, but you can an idea of the approximate size.
Physics
1 answer:
den301095 [7]3 years ago
4 0

Answer:

Explanation:

Normal length of spring = 28.3 cm

stretched length of spring = 38.2 cm

length of extension = 38.2 - 28.3 = 9.9 cm

= 9.9 x 10⁻² m

force applied to stretch = .55 x 9.8 ( mg )

= 5.39 N

Force constant = force applied / extension

= 5.39 / 9.9 x 10⁻²

= .5444 x 10² N /m

= 54.44 N/m

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A 150 kg safe on frictionless casters is being raised at 1.20m to the bed of a truck using planks 4m long. The force needed to p
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Read 2 more answers
A transparent oil with index of refraction 1.28 spills on the surface of water (index of refraction 1.33), producing a maximum o
Ad libitum [116K]

Answer:

The thickness of the oil slick is 1.95\times10^{-7}\ m

Explanation:

Given that,

Index of refraction = 1.28

Wave length = 500 nm

Order m = 1

We need to calculate the thickness of oil slick

Using formula of thickness

2nt= m\lambda

Where, n = Index of refraction

t = thickness

\lambda = wavelength

Put the value into the formula

2\times1.28 \times t=1\times\times500\times10^{-9}

t = \dfrac{1\times\times500\times10^{-9}}{2\times1.28 }

t=1.95\times10^{-7}\ m

Hence, The thickness of the oil slick is 1.95\times10^{-7}\ m

4 0
3 years ago
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