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azamat
3 years ago
13

Please answer my question I am having some trouble, please I would love for some help please and thank you.

Physics
1 answer:
solong [7]3 years ago
8 0

the cliff doesnt matter think dropping  a ball at head level around 6 feet and it has a force of 1 unit but you throw it 12 feet up and it falls with a force of 2 units aslong as the Coriolis effect isnt there so ball 1 has more force and it also has more air time as it goes up a certain distance falls back down from a greater height as it has been thrown straight up

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One mole of iron (6 x 10^23 atoms) has a mass of 56 grams, and its density is 7.87 grams per cubic centimeter, so the center-to-
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Answer:

Explanation:

Given that:

length l = 2.3 m

a = 0.12 cm = 0.12  \times 10^{-2} \ m

x = 1.17 \ cm = 1.17 \times 10^{-2}\ m

m = 149 kg

\delta = 7.87 \ g/cm^3

da = 2.28 \times 10^{-10}\ m

F_{net} = F-mg\\ \\0 = F - mg \\ \\  F = mg \\ \\ k_sx = mg \\ \\

∴

k_s = \dfrac{149(9.8)}{1.17 \times 10^{-2}} \\ \\  k_s = 124803.42  \ N /m

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N_{chain} = \dfrac{(a)^2}{(da)^2} = (\dfrac{a}{da})^2

N_{chain} =  (\dfrac{0.12 \times 10^{-2} }{2.28 \times 10^{-10}})^2

N_{chain} = 2.77 \times 10^{13}

N_{bond} = \dfrac{L}{da} \\ \\  = \dfrac{2.3}{2.28 \times 10^{-10}} \\ \\ N_{bond} = 1.009 \times 10^{10}

\text{Finally; the stiffness of a single interatomic spring is:}

k_{si} =\dfrac{N_{bond}}{N_{chain}}\times k_s

k_{si} =\dfrac{(1.009 \times 10^{10})}{2.77*10^{13}}}\times (124803.42)

\mathbf{k_{si} =45.46 \ N/m}

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2 years ago
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