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:
D) four times as great
Explanation:
The current through a circuit or device is given by the formula:
I = q/t
q = (I)(t)
where,
I = current
q = charge
t = time
<u>FOR AAA CELL</u>:
I = 750 mA = 0.75 A
t = 5 min = (5)(60) sec = 300 sec
q = q₁
Therefore,
q₁ = (0.75 A)(300 sec)
q₁ = 225 Coulomb
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<u>FOR C CELL</u>:
I = 750 mA = 0.75 A
t = 20 min = (20)(60) sec = 1200 sec
q = q₂
Therefore,
q₂ = (0.75 A)(1200 sec)
q₂ = 900 Coulomb
Thus, from the values of charges for both cells, it is clear that, Compared to the total charge transferred by the AAA cell through the bulb, the total charge transferred by the C cell through the bulb is:
<u>D) four times as great</u>
Answer:
b. F1=F2=F3=0 because there is no change in the horizontal motion of the arrows.
Explanation:
The three arrows in the given problem have been shot and it is assumed that air resistance can be neglected. The motions of the arrows are in horizontal direction and as such there is no obvious change in the horizontal motion of the three arrows. Therefore, the three forces are equivalent to one another and each force is equal to zero (i.e. F1=F2=F3=0).
Answer:3.87*10^-4
Explanation:
What is the decrease in mass, delta mass Xe , of the xenon nucleus as a result of this deca
We have been given the wavelength of the gamma ray, find the frequency using c = freq*wavelength.
C=f*lambda
3*10^8=f*3.44*10^-12
F=0.87*10^20 hz
Then with the frequency, find the energy emitted using equation
E=hf E = freq*Plank's constant
E=.87*10^20*6.62*10^-34
E=575.94*10^(-16)
With this energy, convert into MeV from joules.
With the energy in MeV, use E=mc^2 using c^2 = 931.5 MeV/u.
Plugging and computing all necessary numbers gives you
3.87*10^-4 u.
