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Over [174]
3 years ago
13

Find the pressure exerted by a waterbed with dimensions of 2 m x 2 m which is 30 cm thick. (hint: 1000 kg/m3 density of water)

Physics
1 answer:
ArbitrLikvidat [17]3 years ago
4 0

Complete question:

Find the pressure exerted by a waterbed with dimensions of 2 m x 2 m which is 30 cm thick. (hint: use 1000 kg/m³ as density of water)

Answer:

The pressure exerted by the waterbed is 2940 N/m²

Explanation:

Given;

Length of waterbed, L = 2 m

Width of waterbed, W = 2 m

Height of waterbed, H = 30 cm = 0.3 m

density of water, ρ = 1000 kg/m³

Hydrostatic pressure derivation:

Pressure = \frac{Force}{Area} = \frac{mg}{L*W} =\frac{(\rho V)g}{L*W}= \frac{(V) \rho g}{L*W} =\frac{(L*W*H)\rho g}{L*W} = \rho gH

The hydrostatic pressure exerted by the waterbed is directly proportional to the height of the waterbed. Thus, the hydrostatic pressure increases with increase in height of the waterbed.

Hydrostatic Pressure exerted by the waterbed:

P = ρgH

P = 1000 x 9.8 x 0.3

P = 2940 N/m²

Therefore, the pressure exerted by the waterbed is 2940 N/m²

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Answer:

a)  R = 2.5 mi   b)  To return to your case you must walk in the opposite direction or θ = 98º

This is 8º north west

Explanation:

This is a distance exercise with vectors the best way to work these is to decompose the vectors and perform the sum on each axis separately

To use the Cartesian system all angles must be measured from the positive side of the x-axis or the signs of the components must be assigned manually depending on the quadrant where they are.

First vector A = 2 to 20º north west

Measured from the positive x axis is θ = 180 -20 = 160º

We use trigonometry to find the components

     Cos 20 = Aₓ / A

     sin 20 = A_{y} / A

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    A_{y}  = A sin160 = 2 sin160

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    A_{y}  = 0.684 mi

Second vector B = 4 mi 10º west of the south

Angle θ = 270 - 10 = 260º

    cos 2600 = Bₓ / B

    sin 260 = B_{y} / B

    Bₓ = B cos 260

     B_{y}  = B sin 260

    Bₓ = 4 cos 260

     B_{y}  = 4 sin 260

     Bₓ = -0.6946mi

     B_{y}  = - 3,939 mi

Third vector C = 3 mi to 15 north east

     cos 15 = Cₓ / C

     sin15 = C_{y} / C

     Cₓ = C cos 15

     C_{y} = C sin15

     Cₓ = 3 cos 15

    C_{y} = 3 sin 15

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     R = √ (0.3241 2 + (-2.4785) 2)

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b) let's use trigonometry

     Tan θ = y / x

     Tanθ  = -2.4785 / 0.3241

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