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vredina [299]
1 year ago
7

Find the density of a liquid, in lb/ft^3, that exerts a pressure of 0.400 lb/in^2 at a depth of 42.0 in.

Physics
1 answer:
Arturiano [62]1 year ago
6 0

Given :

Pressure, P = 0.4 lb/in².

Depth, h = 42 in.

To Find :

The density of a liquid.

Solution :

Pressure at height h of a liquid of density \rho is given by :

P = \rho g h   ....1)

Here, g = 386.04 in/s²( acceleration due to gravity )

Putting all values in equation 1, we get :

P = \rho g h\\\\0.4 = \rho\times 386.04 \times 42\\\\\rho=\dfrac{0.4}{42\times 386.04}\ lb/in^3\\\\\rho=\dfrac{0.4}{42\times 386.04}\times 12^3\ lb/ft^3\\\\\rho=0.0426\ lb/ft^3

Hence, this is the required solution.

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2 years ago
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A bee wants to fly to a flower located due North of the hive on a windy day. The wind blows from East to West at speed 6.68 m/s.
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Answer:  53.31\° East of North

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Speed of the bee relative to the air:  8.33 m/s

If we graph these speeds (which in fact are velocities because are vectors) in a vector diagram, we will have a right triangle in which the airspeed of the bee (its speed relative to te air) is the hypotense and the two sides of the triangle will be the <u>Speed of the wind from East to West</u> (in the horintal part) and the <u>speed due North relative to the ground</u> (in the vertical part).

Now, we need to find the direction the bee should fly directly to the flower (due North):

sin \theta=\frac{Windspeed-from-East-to-West}{Speed-bee-relative-to-air}

sin \theta=\frac{6.68 m/s}{8.33 m/s}

Clearing \theta:

\theta=sin^{-1} (\frac{6.68 m/s}{8.33 m/s})

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