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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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The spin cycle of a clothes washer extracts the water in clothing by greatly increasing the water's apparent weight so that it i
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The apparent weight of a 1.1 g drop of water is 4.24084 N.

<h3>What is Apparent Weight?</h3>
  • According to physics, an object's perceived weight is a characteristic that describes how heavy it is. When the force of gravity acting on an object is not counterbalanced by a force of equal but opposite normality, the apparent weight of the object will differ from the actual weight of the thing.
  • By definition, an object's weight is equal to the strength of the gravitational force pulling on it. It follows that even a "weightless" astronaut in low Earth orbit, with an apparent weight of zero, has almost the same weight that he would have if he were standing on the ground; this is because the gravitational pull of low Earth orbit and the ground are nearly equal.

Solution:

N = Speed of rotation = 1250 rpm

D = Diameter = 45 cm

r = Radius = 22.5 cm

M = Mass of drop = 1.1 g

Angular speed of the water = \omega  = \frac{2\pi N}{60}

\omega  = \frac{2\pi \times 1250}{60}

\omega  = 130.89 rad/s

Apparent weight is given by

W _a = M\omega^{2}R

W_a = 1.1 \times 10^-^3\times (130.89)^2\times 0.225

W_a = 4.24084 N

Know more about Apparent weight brainly.com/question/14323035

#SPJ4

Question:

The spin cycle of a clothes washer extracts the water in clothing by greatly increasing the water's apparent weight so that it is efficiently squeezed through the clothes and out the holes in the drum. In a top loader's spin cycle, the 45-cm-diameter drum spins at 1250 rpm around a vertical axis. What is the apparent weight of a 1.1 g drop of water?

7 0
2 years ago
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