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3 years ago

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A particle, originally at rest and located at point (3 ft, 2 ft, 5 ft), is subjected to an acceleration a = {6ti + 12t^2k} ft/s^

2. Determine the particle’s position (x, y, z) when t = 2 s.

1 answer:

Sindrei [870]3 years ago

4 0

Answer:

The particle's position at t = 2 is

r = (11ft, 2ft, 21ft) = (11, 2, 21) ft

Explanation:

r₀ = (3, 2, 5) ft

a = (6t, 0, 12t²) ft/s²

a = dv/dt

dv/dt = (6tî + 0j + 12t²ķ)

dv = (6tî + 0j + 12t²ķ) st

Integrating the left hand side from 0 to v (the particle was originally at rest) and the right hand side from 0 to t,

We obtain,

v = (3t²î + 0j + 4t³ķ) ft/s

v = dr/dt

dr/dt = (3t²î + 0j + 4t³ķ)

dr = (3t²î + 0j + 4t³ķ) st

Integrating the left hand side from r₀ (the original position) to r and the right hand side from 0 to t,

r - r₀ = (t³î + 0j + t⁴ķ) ft

r = (t³î + 0j + t⁴ķ) + r₀

r = (t³î + 0j + t⁴ķ) + (3î + 2j + 5ķ)

At t=2s, t³ = 8 and t⁴ = 16

r = (8î + 0j + 16ķ) + (3î + 2j + 5ķ)

r = (11î + 2j + 21ķ) ft

r = (11ft, 2ft, 21ft)

Hope this helps!

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Initially when 1000.00 mL of water at 10oC are poured into a glass cylinder, the height of the water column is 1000.00 mm. The w

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Answer:

$\mathbf{h_2 =1021.9 \ mm}$

Explanation:

Given that :

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$V_1 = A_1 *h_1$

The area of the water before heating is:

$A_1 = \dfrac{V_1}{h_1}$

$A_1 = \dfrac{1000000}{1000}$

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$A_2 = A_1 (1 + \Delta t \alpha )^2$

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Hence the depth of the heated hot water is $\mathbf{h_2 =1021.9 \ mm}$

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Solving the equations (i) and (ii),

$\frac{H_{f_1}}{H_{f_2}}=\frac{\frac{8flQ_1^2}{\pi^2 gd_1^5}}{\frac{8flQ_2^2}{\pi^2 gd_2^5}}$

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Therefore,

$\frac{H_{f_1}}{H_{f_2}}=\frac{1}{10}$

or $H_{f_2}=10 \ H_{f_1}$

Thus the answer is option A). 10

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3 years ago