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3 years ago

HELP PLEASE! write the equation for calculating relative velocity in one dimensional.

Physics

1 answer:

RoseWind [281]3 years ago

6 0

Anwer:

Key Equations

Position Vector: `n r (t) = (t)^j+y

Total acceleration: `n a = `n aC

+ `n aT

Position Vector in frame S is the position Vector in frame S` plus the vector from the origin of S to the origin of S`: `n rPS = `n rPS`

+ `n rS`S

Relative velocity equation connecting two references frame: `n vPS = `n vPS`

+ `n vS`S

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Answer:

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Explanation:

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3 years ago

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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-

choli [55]

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

$\delta = 7.87 \ g/cm^3$

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

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

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

$N_{chain} = \dfrac{A_{wire}}{A_{atom}} = \dfrac{A_w}{da^2}$

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

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

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

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

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

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3 years ago

11. An object moves in circular path with constant speed

Temka [501]

Answer:

B. Is its acceleration constant

Explanation:

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beks73 [17]

Mass = density • volume

so (12)(920)(kg/m^3•m^3)

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3 years ago

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zvonat [6]

The best and most correct answer among the choices provided by the question is the first choice, larger.

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