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2 years ago

Is velocity ratio of a machine affected by applying oil on it?Explain with reason.

1 answer:

6 0

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.

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A scuba diver has his lungs filled to half capacity (3 liters) when 10 m below the surface. If the diver holds his breath while

rjkz [21]

To solve this problem it is necessary to apply Boyle's law in which it is specified that

$P_1V_1 =P_2 V_2$

Where,

$P_1$ and $V_1$ are the initial pressure and volume values

$P_2$ and $V_2$ are the final pressure volume values

The final pressure here is the atmosphere, then

$P_2 = 101325 \approx 1*10^5Pa$

$h = 10m$

$\rho_w = 1000kg/m^3$

$V_1 = 3.0L$

Pressure at the water is given by,

$P_1 = P_2 -\rho gh$

$P_1 = 1*10^5 +1000*9.8*10 =198000Pa$

Using Boyle equation we have,

$V_2 = \frac{P_1V_1}{P_2}$

$V_2 = \frac{198000*3*10^5}{10^5}$

$V_2 = 5.9L$

Therefore the volume of the lungs at the surface is 5.9L

4 0

1 year ago

Classes are canceled due to snow, so you take advantage of the extra time to conduct some physics experiments. You fasten a larg

horrorfan [7]

Answer:

The value is $t_1 = 9 \ s$

Explanation:

Generally the velocity attained by the sled after t = 3.10 s is mathematically evaluated using the kinematic equation as follows

$v = u + at$

Here u = 0 \ m/s

a = 13.5 $m/s^2$

$v = 0 + 13.5 * 3.10$

=> $v = 41.85 \ m/s$

The is distance it covers at this time is

$s = u * t + \frac{1}{2} a * t^2$

=> $s = + \frac{1}{2} * 13.5 * 3.10^2$

=> $s =64.87$

Now when sled stops its the final velocity is $v_f = 0 m/s$ while the initial velocity will be the velocity after its acceleration i.e $v = 41.85 \ m/s$