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Zinaida [17]
2 years ago
10

How do sinkholes occur? Overburden falls into the space left behind when bedrock is dissolved. Bedrock falls under the pressure

of the overburden. Overburden is dissolved by water, wind, and acids. Bedrock collapses under the weight of buildings/objects on the surface.
Physics
1 answer:
BabaBlast [244]2 years ago
6 0

Answer: over burden is dissolved by water wind and acids

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Two copper rods are separated by a small gap at B. Rod AB has a diameter of 200mm and rod BC has a diameter of 150mm. Find the f
gulaghasi [49]

Answer:

A)3.8196 * 10^{9} Newtons        B) 2.153 * 10^{9} Newtons

Explanation:

Force = Pressure * Area.

Pressure is given as  120GPa

Area = Cross Sectional Area of Rod = Area Of A circle = πR^{2}.

Area Expansivity β =    \frac{Change in Area}{Original Area * Temperature Rise}

Area Expansivity β = 2α

α =  16.9x10-6/ ° C.

Radius of Rod AB = Half 0f Diameter, 200mm = \frac{0.2m}{2} = 0.1 m

Radius of Rod BC = Half Of Diameter, 150mm = \frac{0.15}{2} = 0.075 m.

Note: To convert from millimeter to meter you divide by 1000.

Area of Rod AB = π * 0.1^{2} =  0.0314m^{2}

Area of Rod BC = π * 0.075^{2} = 0.0177m^{2}.

Change in Area = 2α * Original Area * Temperature Rise. Therefore for

Rod AB = 2 * 16.9*10^{-6} * 0.0314 * 30 = 3.183 * 10^{-5}m^{2}.

For Rod BC we have = 2 * 16.9*10^{-6} * 0.0177 * 30 = 1.794∈-5m^{2}.

The forces on each rods will be given by.

Force AB = Pressure * Area  of Rod AB

                 = 120GPa * 3.183 * 10^{-5} gives 3.8196 * 10^{9} Newtons.

Force BC = Pressure * Area of Rod BC

                = 120GPa * 1.794 * 10^{-5} gives 2.153 * 10^{9} Newtons

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A ball that has a mass of 0.25 kg spins in a circle at the end of a 1.6 m rope. the ball moves at a tangential speed of 12.2 m/s
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The centripetal force acting on the ball will be 23.26 N.The direction of the centripetal force is always in the path of the center of the course.

<h3>What is centripetal force?</h3>

The force needed to move a body in a curved way is understood as centripetal force. This is a force that can be sensed from both the fixed frame and the spinning body's frame of concern.

The given data in the problem is;

m is the mass of A ball = 0.25 kg

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\rm F_C is the centripetal force acting on the ball

The centripetal force is found as;

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Hence the centripetal force acting on the ball will be 23.26 N.

To learn more about the centripetal force refer to the link;

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True.

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