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.
Answer: (a) The gravitational force on the object at the North Pole of Neptune is 51.7N
(b) The apparent weight of the object at Neptune's equator is 50.4N
Explanation: Please see the attachments below
Answer:
Jari
Explanation:
The question requires to know who is traveling faster. This is done by comparing the gradients. The steeper the slope (high gradient), the faster the speed and vice versa.
From Jari's line, the starting point is (0, 0) and another point is (6, 7)
The gradient being change in y to change in x
Change in y=7-0=7
Change in x=6-0=6
Slope is 7/6
For Jade, first point is (0, 10) then another point is (6, 16)
Change in y=16-10=6
Change in x=6-0=6
Slope is 6/6=1
Clearly, 7/6 is greater than 6/6 or 1 hence Jari is faster than Jade
The unit 'mb' means millibar which is equivalent to 1/1000 of 1 bar. To convert the units from bar to atmospheres (atm) and to inches Hg (inHg), we need to know the conversion factors.
a.) 1 atm = 1.01325 bar
0.92 mb(1 bar/1000 mbar)(1 atm/1.01325 bar) =<em> 9.08×10⁻⁴ atm</em>
b.) 1 bar = 29.53 inHg
0.92 mb(1 bar/1000 mbar)(29.53 inHg/1 bar) =<em> 0.027 inHg</em>
We can first calculate the net force using the given information.
By Newton's second law, F(net) = ma:
F(net) = 25 * 4.3 = 107.5
We can now calculate the frictional force, f, which is working against the applied force, F(app) (this is why the net force is a bit lower):
f = F(net) - F(app) = 150 - 107.5 = 42.5 N
Now we can calculate the coefficient of friction, u, using the normal force, F(N):
f = uF(n) --> u = f/F(N)
u = 42.5/[25(9.8)]
u = 0.17
