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Stolb23 [73]
3 years ago
13

How can the density of irregularly shaped objects can be calculated?

Physics
1 answer:
Aleks04 [339]3 years ago
6 0
1.Use the balance to find the mass of the object. Record the value on the "Density Data Chart."
2.Pour water into a graduated cylinder up to an easily-read value, such as 50 milliliters and record the number.
3.Drop the object into the cylinder and record the new value in millimeters.
4.The difference between the two numbers is the object's volume. Remember that 1 milliliter is equal to 1 cubic centimeter. Record the volume on the data chart.
5.Compute the density of the object by dividing the mass value by the volume value. Record the density on the data chart.
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3 years ago
A rectangular tank is filled to a depth of 10m with freshwater and open to air at atmospheric pressure.
charle [14.2K]

Answer:

<em>1.</em> <em>39068.07 N</em>

<em>2. 19534.036 N</em>

Explanation:

depth of water h = 10 m

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density of water p = 1000 kg/m^3

acceleration due to gravity g = 9.81 m/s^2

pressure due to depth of water = pgh

P = 1000 x 9.81 x 10 = 98100 Pa

total pressure on the bottom of the tank is Patm + p = 101325 + 98100 = 199425 Pa

Left plug has diameter = 50 cm = 0.5 m

radius = 0.5/2 = 0.25 m

height = 1 cm = 0.01 m

<em>height below tank surface = 10 - 0.01 = 9.99</em>

pressure at this depth =  1000 x 9.81 x 9.99 = 98001.9 Pa

<em>total pressure = Patm + P = 101325 + 98001.9 = 199326.9 Pa</em>

surface area of plug = πr^{2} = 3.142 x 0.25^{2} = 0.196 m^{2}

<em>force required to lift left plug = pressure x area</em>

F =  199326.9 x 0.196 = <em>39068.07 N</em>

<em>The right side is a hemisphere with the same diameter, therefore surface area is half of the left plug</em>

A = 0.196/2 = 0.098 m^{2}

force F required to lift right plug =  199326.9  x 0.098 =<em> 19534.036 N</em>

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Explanation: I hoped That Helped !!

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