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liubo4ka [24]
3 years ago
7

How is buoyancy used in the real world?​

Physics
2 answers:
JulsSmile [24]3 years ago
7 0

Answer:

to mack boats float

Explanation:

without it the boats will sink to the bottom

seraphim [82]3 years ago
4 0

-- Buoyancy is used to bring crude oil from Saudi Arabi to refineries in the US. The oil is loaded into tanks that are part of supertanker ships. Buoyancy is created by their ship-shape, so they float on water, and they can be dragged across the ocean on top of the water.

-- Buoyancy is used for a cheap thrill on the weekend.  People drive out in the country and find a place where they can buy a ride under a hot-air balloon.  The balloon is filled with hot air from a propane burner under its opening, and then it rises up off the ground because of its buoyancy in cool air.

-- Buoyancy is used cleverly by factories, to get rid of their gaseous and particulate wastes.  The wastes are heated, and then blown into tall "smokestacks" connected to the factory.  Then they rise because of the buoyancy created by hot stuff in cool air.  When they leave the top of the smokestack, they keep rising for a while.  Then, weather systems blow them away from the factory, over into other people's neighborhoods, where they finally sink to the ground in places where nobody knows where they came from.

-- Buoyancy is used in teapots, coffee makers, and lava lamps.  A heat source is placed under the bottom of the container. Heat is conducted through the bottom of the container, and it heats the fluid that's in contact with the inside of the container on the bottom. The hot fluid rises to the top of the container, because of the buoyancy created by hot stuff surrounded by cool stuff.  At the same time, cool stuff flows in to take its place at the bottom.  

In pots and pans and tanks where buoyancy is used this way, the motion of the fluids up and down and around is called "convection".

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Suppose you are drinking root beer from a conical paper cup. The cup has a diameter of 10 centimeters and a depth of 13 centimet
sveta [45]

Answer:

The level of the root beer is dropping at a rate of 0.08603 cm/s.

Explanation:

The volume of the cone is :

V=\frac {1}{3}\times \pi\times r^2\times h

Where, V is the volume of the cone

r is the radius of the cone

h is the height of the cone

The ratio of the radius and the height remains constant in overall the cone.

Thus, given that, r = d / 2 = 10 / 2 cm = 5 cm

h = 13 cm

r / h = 5 / 13

r = {5 / 13} h

V=\frac {1}{3}\times \frac {22}{7}\times ({{{\frac {5}{13}\times h}}})^2\times h

V=\frac {550}{3549}\times h^3

Also differentiating the expression of volume w.r.t. time as:

\frac {dV}{dt}=\frac {550}{3549}\times 3\times h^2\times \frac {dh}{dt}

Given: \frac {dV}{dt} = -4 cm³/sec (negative sign to show leaving)

h = 10 cm

So,

-4=\frac{550}{3549}\times 3\times {10}^2\times \frac {dh}{dt}

\frac{55000}{1183}\times \frac {dh}{dt}=-4

\frac {dh}{dt}=-0.08603\ cm/s

<u>The level of the root beer is dropping at a rate of 0.08603 cm/s.</u>

3 0
3 years ago
Which statement is NOT true?
Nuetrik [128]

Answer:

light doesn't need a medium through which to travel because the speed of light is experimentally constant

4 0
3 years ago
The process of burning fuel is called?
user100 [1]
The process of burning fuel is Combustion
5 0
3 years ago
Read 2 more answers
A pebble is released from rest at a certain height and falls freely, reaching an impact speed of 6 m/s at the floor. Next, the p
Anna71 [15]

Answer:

Explanation:

Let h be the height .

initial velocity in first case u = 0

final velocity v = 6 m /s

acceleration due to gravity g = 9.8 m /s²

v² = u² + 2 g h

6² = 0 + 2 x 9.8 x h

h = 1.837 m .

For second case u = 3 m /s

v² = u² + 2 gh

= 3² + 2 x 1.837 x 9.8

= 9 + 36

= 45 m

v = 6.7 m /s

8 0
3 years ago
A large grinding wheel in the shape of a solid cylinder of radius 0.330 m is free to rotate on a frictionless, vertical axle. a
USPshnik [31]
Refer to the diagram shown below.

Let I = the moment of inertia of the wheel.
α = 0.81 rad/s², the angular acceleration
r = 0.33 m, the radius of the weel
F = 260 N, the applied tangential force

The applied torque is
T = F*r
   = (260 N)*(0.33 m)
   = 85.8 N-m

By definition,
T = I*α

Therefore,
I = T/α
  = (85.8 N-m)/(0.81 rad/s²)
  = 105.93 kg-m²

Answer: 105.93 kg-m²

6 0
3 years ago
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