Factors affecting friction
The intensity of friction depends on following factors: i) The area involved in friction. ii) The pressure applied on the surfaces. Force = Pressure ´ Area Frictional force will increase, if the area of contact will increase or if pressure applied on the surface increased.
Methods to reduce friction
i) Polish the contact surface. ii) Put oil or grease so that it fills in the small gaps of the flat parts. iii) Use ball bearings to reduce area of contact between rotating parts.
Lubrication
Following methods can be used to reduce friction: Oil is either thin or viscous. It depends upon SAE No. of oil. (SAE means Society of Automotive Engineers). If we use very viscous oil, it does not reach all the parts. Very thin oil will flows away easily and gets wasted. Grease is used in such cases. It is generally used around ball-bearing. Normal grease or oil is never used where there is high pressure, high temperature and high speed. Special lubricants are used in such cases. In cold season the oil becomes thick and in hot season it becomes thin. Therefore selection of lubrication also depends on the season. It is always advisable to refer operating manual of the equipment before selecting the lubricant.
Donna's doctor must have diagnosed her with muscle cramps. With the given symptoms above, it is likely that she is experiencing muscle cramps. Muscle cramps most likely happen when the individual experiencing it is lack of fluid intake, causing the muscles to tighten causing pain and for the muscle to contract involuntarily.
The density of the substance is the ratio of its mass over the space it occupies. In mathematical equation, this can be expressed as,
ρ = m / v
where ρ is density, m is mass, and v is volume.
Substituting the known values from the given,
ρ = (45 g) / (8 cm³)
ρ = 5.625 g/cm³
<em>ANSWER: 5.625 g/cm³</em>
If we are talking on the force being exerted by a segment of a rope of lenght R on the right on a point M which is being also pulled from the Left by a segment of rope R as shown in the figure attached. Then we invoke Newton's Third Law:
"Any force exerted by an object (in this case a segment of the rope) also suffers a equal and opposite force".
If we pick

whis is the tension exerted by the right segment then the left segment will also exert an equal and opposite force so we have that
Answer:
Diameter of the cylinder will be 
Explanation:
We have given young's modulus of steel
Change in length 
Length of rod 
Load F = 11100 KN
Strain is given by 
We know that young's modulus 
So 

We know that stress 
So 

So 