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.
As the temperature changes and their masses are the same, heat lost by the balls is directly proportional to their specific heat values. The heat lost by the aluminum ball is higher implies aluminum has higher specific heat.
Answer:
I1 = 0.772 A
Explanation:
<u>Given</u>: R1 = 5.0 ohm, R2 = 9.0 ohm, R3 = 4.0 ohm, V = 6.0 Volts
<u>To find</u>: current I = ? A
<u>Solution: </u>
Ohm's law V= I R
⇒ I = V / R
In order to find R (total) we first find R (p) fro parallel combination. so
1 / R (p) = 1 / R1 + 1/ R2 ∴(P) stand for parallel
R (p) = R1R2 / ( R1 + R2)
R (p) = (5.0 × 9.0) / (5.0 + 9.0)
R (p) = 3.214 ohm
Now R (total) = R (p) + R3 (as R3 is connected in series)
R (total) = 3.214 ohm + 4.0 Ohm
R (total) = 7.214 ohm
now I (total) = 7.214 ohm / 6.0 Volts
I (total) = 1.202 A
This the total current supplied by 6 volts battery.
as voltage drop across R (p) = V = R (p) × I (total)
V (p) = 3.214 ohm × 1.202 A = 3.864 volts
Now current through 5 ohms resister is I1 = V (P) / R1
I1 = 3.864 volts / 5 ohm
I1 = 0.772 A
Answer:
a. be sure to hold expansion cards by the edge connectors
Explanation:
Removal of loose jewelry is a good safety practice. Also not touching a microchip with a magnetized screwdriver is also a good practice.
But holding expansion cards by the edge connectors is not a good practice, so it is the odd one in the question. Therefore answer option a provides the correct and best answer to the question
Hope this is helpful <span>Weightlessness
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