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kumpel [21]
2 years ago
9

What is meant by rusting​

Physics
2 answers:
Aneli [31]2 years ago
8 0

Answer:

<em>The</em><em> </em><em>formation</em><em> </em><em>of</em><em> </em><em>reddish</em><em> </em><em>brown</em><em> </em><em>ferric</em><em> </em><em>oxide</em><em> </em><em>on</em><em> </em><em>iron</em><em> </em><em>by</em><em> </em><em>low</em><em> </em><em>temperature</em><em> </em><em>oxidation</em><em> </em><em>in</em><em> </em><em>the</em><em> </em><em>presence</em><em> </em><em>of</em><em> </em><em>water</em><em> </em><em>.</em><em> </em>

<em>make</em><em> </em><em>me</em><em> </em><em>as</em><em> </em><em>brilliant</em><em> </em><em>answer</em><em> </em>

Setler79 [48]2 years ago
3 0

The process of formation of a reddish-brown substance on the surface of the iron objects in the presence of flaky moisture and air is called rusting.

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Citrus2011 [14]

The maximum speed is 10.4 m/s

Explanation:

For a body in uniform circular motion, the centripetal acceleration is given by:

a=\frac{v^2}{r}

where

v is the linear speed

r is the radius of the circular path

In this problem, we have the following data:

- The maximum centripetal acceleration must be

a=1.1 g

where g=9.8 m/s^2 is the acceleration of gravity. Substituting,

a=(1.1)(9.8)=10.8 m/s^2

- The radius of the turn is

r = 10 m

Therefore, we can re-arrange the equation to solve for v, to find the maximum speed the ride can go at:

v=\sqrt{ar}=\sqrt{(10.8)(10)}=10.4 m/s

Learn more about centripetal acceleration:

brainly.com/question/2562955

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3 years ago
The coefficient of the restitution of an object is defined as the ratio of its outgoing to incoming speed when the object collid
IgorLugansk [536]

Answer:

48.16 %

Explanation:

coefficient of restitution = 0.72

let the incoming speed be = u

let the outgoing speed be = v

kinetic energy = 0.5 x mass x x velocity^{2}

  • incoming kinetic energy = 0.5 x m x x u^{2}

     

  •  coefficient of restitution =\frac{v}{u}

       0.72 =\frac{v}{u}

       v = 0.72u

        therefore the outgoing kinetic energy = 0.5 x m x (0.72u)^{2}

        outgoing kinetic energy = 0.5 x m x 0.5184 x u^{2}

        outgoing kinetic energy = 0.5184 (0.5 x m x x u^{2})

recall that 0.5 x m x x u^{2} is our incoming kinetic energy, therefore

outgoing kinetic energy = 0.5184 x (incoming kinetic energy)

from the above we can see that the outgoing kinetic energy is 51.84 % of the incoming kinetic energy.

The energy lost would be 100 - 51.84 = 48.16 %

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