An aquarium is made of glass that has the density of 2500
kg/m³ and having
the thickness of 4mm. the aquarium has the length of 1 meter, the breadth of 40
cm and the height of 25 cm. the aquarium is placed on a wooden table. Somebody
then filled it with water. The volume of the water is 60 liter and its density
1000 kg/m³. If we are using the gravitational acceleration of 9,8
m/s² ; decide :
a. the height of the water in the
aquarium
b. the weight of the aquarium when it is
still empty
c. the weight of the aquarium when It is
already filled with water
d. the pressure caused by the aquarium
with water to its table
e. if there’s a turtle living at the
bottom of the aquarium, how big is the hydrostatic pressure received by the
turtle
yes, it is the dimension of its interior. it has no lid
Answers
Answered by
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a.
height of water in the aquarium =
= volume of water filled / area of bottom of aquarium
= 60, 000 cm³ / (100 cm * 40 cm) = 15 cm
If the length and breadth given are the dimensions outside (external), then
height = 60, 000 cm³ / [ (100 - 2 * 0.4) cm * (40 - 2 * 0.4) cm ]
= 15.43 cm
b) if given dimensions are internal dimensions of aquarium. then:
(there is no top lid)
external Volume of aquarium = (100 + 0.4*2)(40+0.4*2)(25+0.4*1) cm³
= 1, 04, 461.056 cm³
Internal volume = 100 * 40 * 25 = 1,00,000 cm³
Volume of the glass material in the aquarium = 4, 461.056 cm³
Weight of the aquarium = 4, 461.056 cm³ * 2.5 g/cm³ * 980 cm/sec²
= 109, 29, 587.2 dynes = 109.30 Newtons
c)
Weight of 60 liters of water = 60 kg
Total weight of aquarium with water = 169.30 newtons
d)
Pressure = Force on the table / Area of exterior at the bottom of aquarium
= 169.30 Newtons / (1.08 m * 0.48 m) = 326.581 Pa or N/m²
Here we are not considering the weight of atmosphere above aquarium.
e)
Hydrostatic pressure on the turtle = weight of water column at the bottom of aquarium
= 15.43 cm of water column * 1 gm/cm³ * 980 cm/sec²
= 15121.4 dyne = 0.151214 Pa
If the standard atmospheric pressure (of air above water level ) is to be added in addition to the hydrostatic pressure then, 1 atm is to be added.
height of water in the aquarium =
= volume of water filled / area of bottom of aquarium
= 60, 000 cm³ / (100 cm * 40 cm) = 15 cm
If the length and breadth given are the dimensions outside (external), then
height = 60, 000 cm³ / [ (100 - 2 * 0.4) cm * (40 - 2 * 0.4) cm ]
= 15.43 cm
b) if given dimensions are internal dimensions of aquarium. then:
(there is no top lid)
external Volume of aquarium = (100 + 0.4*2)(40+0.4*2)(25+0.4*1) cm³
= 1, 04, 461.056 cm³
Internal volume = 100 * 40 * 25 = 1,00,000 cm³
Volume of the glass material in the aquarium = 4, 461.056 cm³
Weight of the aquarium = 4, 461.056 cm³ * 2.5 g/cm³ * 980 cm/sec²
= 109, 29, 587.2 dynes = 109.30 Newtons
c)
Weight of 60 liters of water = 60 kg
Total weight of aquarium with water = 169.30 newtons
d)
Pressure = Force on the table / Area of exterior at the bottom of aquarium
= 169.30 Newtons / (1.08 m * 0.48 m) = 326.581 Pa or N/m²
Here we are not considering the weight of atmosphere above aquarium.
e)
Hydrostatic pressure on the turtle = weight of water column at the bottom of aquarium
= 15.43 cm of water column * 1 gm/cm³ * 980 cm/sec²
= 15121.4 dyne = 0.151214 Pa
If the standard atmospheric pressure (of air above water level ) is to be added in addition to the hydrostatic pressure then, 1 atm is to be added.
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And does the aquarium contain a lid on top of it?