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3 years ago

An oil tank has a base of area 2.5 m2 and is filled with oil to a depth of 1.2 m. The density of the

Physics

1 answer:

asambeis [7]3 years ago

3 0

Answer:

D. F = 24000[N]

Explanation:

To be able to calculate the force we must first find the pressure at the bottom of the tank. By means of the following equation:

$P=Ro*g*h$

where:

Ro = density of the liquid = 800 [kg/m³]

g = gravity acceleration = 9.81 [m/s²]

h = elevation = 1.2 [m]

$P=800*9.81*1.2\\P=9417.6[Pa]$

Pressure is defined as the relationship between Force on the area.

$P=F/A\\F=9417.6*2.5\\F = 23544 [N] = 24000[N]$

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kap26 [50]

$W = K[ {\frac{1}{3} - \frac{1}{\sqrt{38} } ]$

<u>Explanation:</u>

The parametric representation of a line segment joining the points (a,b,c) and (l,m,n) is

r(t) = (1-t) . (a,b,c) + t . (l, m, n) where t ∈ |0, 1|

So, the parametric representation of a line segment joining the points (3,0,0) and (3,2,5) is

r(t) = (1 - t) . (3,0,0) + t . (3,2,5) where t ∈ |0, 1|

r(t) = (3(1 - t), 0, 0) + (3t, 2t, 5t) where t ∈ |0, 1|

r(t) = (3, 2t, 5t)

Given:

$F(x, y, z) = \frac{Kr}{|r|^3} \\\\F(x, y, z) = \frac{K}{(x^2 + y^2 + z^2)^3^/^2} (x, y, z)\\\\F(r(t)) = \frac{K}{(3^2 + (2t)^2 + (5t)^2)^3^/^2} (3, 2t, 5t)\\\\F(r(t)) = \frac{K}{(9 + 29t^2)^3^/^2} (3, 2t, 5t)$

dr = (0, 2, 5) dt

$Work = \int\limits^1_0 {F} \, dr$

$W = \int\limits^1_0 {\frac{K}{(9 + 29t^2)^3^/^2} } (3, 2t, 5t) . (0, 2, 5)\, dt\\ \\ = \int\limits^1_0 {\frac{K ( 4t + 25t)}{(9 + 29t^2)^3^/^2} } \, dt\\\\\\$

$W = \frac{1}{2}\int\limits^1_0 {\frac{K(29t)}{(9 + 29t^2)^3^/^2} } \, dt \\ \\$

Substitute = 9 + 29t² = u, 92tdt = du

Limit changes from 0→1 to 9 → 38

$W = \frac{K}{2} \int\limits^3_9 {\frac{du}{u^3^/^2} } \,\\\\$

On solving this, we get:

$W = K[ {\frac{1}{3} - \frac{1}{\sqrt{38} } ]$

Therefore, work done is $W = K[ {\frac{1}{3} - \frac{1}{\sqrt{38} } ]$

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3 years ago