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2 years ago

Cat is stuck in 10 foot hole how do i get it unstuck

Engineering

2 answers:

Semenov [28]2 years ago

7 0

Answer:

you fill the hole so there is no more cat

sweet [91]2 years ago

7 0

Answer:

I don't even know if you're serious but... you call the fire department and tell them about the issue and they get it unstuck

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At 45° latitude, the gravitational acceleration as a function of elevation z above sea level is given by g = a − bz , where a =

Ahat [919]

Answer:

8861.75 m approximately 8862 m

Explanation:

We need to remember Newton's 2nd Law which says that the force experienced by an object is proportional to his acceleration and that the constant of proportionality between those two vectors correspond to the mass of the object.

$F=ma$ for the weight of an object (which is a force) we have that the acceleration experienced by that object is equal to the gravitational acceleration, obtaining that $W = mg$

For simplicity we work with $g =9.807$ $\frac{m}{s^{2}}$ despiting the effect of the height above sea level. In this problem, we've been asked by the height above sea level that makes the weight of an object 0.30% more lighter.

In accord with the formula $g = a-bz$ the "normal" or "standard" weight of an object is given by $W = mg = ma$ when $z = 0$ , so we need to find the value of $z$ that makes $W = m(a-bz) = 0.997ma$ meaning that the original weight decrease by a 0.30%, so now we operate...

$m(a-bz) = 0.997ma$ now we group like terms on the same sides $ma(1-0.997) = mbz$ we cancel equal tems on both sides and obtain that $z = \frac{a}{b} (0.003) = \frac{9.807 \frac{m}{s^{2} } }{3.32*10^{-6} s^{-2} } (0.003) = 8861.75 m$

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3 years ago

An aluminum alloy tube with an outside diameter of 3.50 in. and a wall thickness of 0.30 in. is used as a 14 ft long column. Ass

slega [8]

Answer:

slenderness ratio = 147.8

buckling load = 13.62 kips

Explanation:

Given data:

outside diameter is 3.50 inc

wall thickness 0.30 inc

length of column is 14 ft

E = 10,000 ksi

moment of inertia $= \frac{\pi}{64 (D_O^2 -D_i^2)}$

$I = \frac{\pi}{64}(3.5^2 -2.9^2) = 3.894 in^4$

Area $= \frac{\pi}{4} (3.5^2 -2.9^2) = 3.015 in^2$

$radius = \sqrt{\frac{I}{A}}$

$r = \sqrt{\frac{3.894}{3.015}$

r = 1.136 in

slenderness ratio $= \frac{L}{r}$

$= \frac{14 *12}{1.136} = 147.8$

buckling load $= P_cr = \frac{\pi^2 EI}}{l^2}$

$P_{cr} = \frac{\pi^2 *10,000*3.844}{( 14\times 12)^2}$

$P_{cr} = 13.62 kips$

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3 years ago

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aleksley [76]

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STatiana [176]

15x -/c/ fb is the answer

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liq [111]

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Explanation:

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