The derived frequency equals 2.63 Hz. Explanation: For an object weighing 8.0 kg with a spring stretching 3.6 cm, calculations involving the spring constant and oscillation frequency lead to this specific oscillation rate.
Answer:
0.018 J
Explanation:
The work required to bring the charge from infinity to the point P is equal to the change in its electric potential energy. This can be expressed as
where
represents the charge's magnitude
and 
signifies the potential difference between point P and infinity.
After substituting into the formula, we arrive at
Factors influencing friction
The magnitude of friction is contingent on the following elements: i) The surface area in contact. ii) The applied pressure on the surfaces. Force is determined by Pressure multiplied by Area; thus, if the contact area increases or if the pressure applied rises, the frictional force will also escalate.
Methods for reducing friction
i) Smooth the contact surface. ii) Apply oil or grease to fill small gaps in flat surfaces. iii) Use ball bearings to minimize contact area among rotating components.
Lubrication
To minimize friction, various methods may be employed: Oil can be either thin or viscous, which depends on its SAE number (SAE indicating Society of Automotive Engineers). Highly viscous oils may not reach all components effectively. In contrast, very thin oils may drain away quickly, resulting in wastage. Grease is preferable in such situations, particularly around ball-bearings. Regular grease or oil should not be utilized under high speed, high pressure, and high temperature conditions—specialized lubricants are required then. The consistency of oil varies with temperature; it thickens in the cold and thins in the heat. Therefore, the choice of lubricant should be seasonally appropriate, and it's always wise to consult the equipment's operating manual prior to making a selection.[[TAG_11]]
Answer:
The value of Y is -10°.
Explanation:
For scale X, the freezing point is 40° and the boiling point is 120°.
The gap between the two endpoints for scale X = 120 - 40 = 80
For scale Y, the ice point and steam point are -30° and 130° correspondingly.
The difference between these two points for scale Y = 130 - (-30) = 160
Comparing both scales:
One unit in scale X is x
One unit in scale Y is y
Scale X consists of 80 divisions, while scale Y has 160
80x = 160y
x = 2y
50° on scale X equals 10x plus the freezing point of scale X
10 divisions in scale Y correspond to 20y
The reading on scale Y = the ice point of Y + 20y
= -30° + 20°
= -10°
