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:
v_avg = 2.9 cm/s
Explanation:
The average velocity of the object is the sum of the distance of all its trajectories divided the time:

x_all is the total distance traveled by the object. In this case you have that the object traveled in the first trajectory 165cm-15cm = 150cm, and in the second one, 165cm - 25cm = 140cm
Then, x_all = 150cm + 140cm = 290cm
The average velocity is, for t = 100s

hence, the average velocity of the object in the total trajectory traveled is 2.9 cm/s
Answer:
-40 kJ
80 kJ
Explanation:
Work is equal to the area under the pressure vs volume graph.
W = ∫ᵥ₁ᵛ² P dV
2.27) Pressure and volume are linearly related. When we graph P vs V, the area under the line is a trapezoid. So the work is:
W = ½ (P₁ + P₂) (V₂ − V₁)
W = ½ (100 kPa + 300 kPa) (0.1 m³ − 0.3 m³)
W = -40 kJ
2.29) Pressure and volume are inversely proportional:
pV = k
The initial pressure and volume are 500 kPa and 0.1 m³. So the constant is:
(500) (0.1) = k
k = 50
The final pressure is 100 kPa. So the final volume is:
(100) V = 50
V = 0.5
The work is therefore:
W = ∫ᵥ₁ᵛ² P dV
W = ∫₀₁⁰⁵ (50/V) dV
W = 50 ln(V) |₀₁⁰⁵
W = 50 (ln 0.5 − ln 0.1)
W ≈ 80 kJ
<h2><u>Answer:</u></h2>
The simulation kept track of the variables and automatically recorded data on object displacement, velocity, and momentum. If the trials were run on a real track with real gliders, using stopwatches and meter sticks for measurement, the data compared by the following statements:
1. (There would be variables that would be hard to control, leading to less reliable data.)
3. (Meter sticks may lack precision or may be read incorrectly.)
4. (Real glider data may vary since real collisions may involve loss of energy.)
5. (Human error in recording or plotting the data could be a factor.)
Answer:
The magnitude of the magnetic force exerted on the moving charge by the current in the wire is 2.18 x
N
The direction of the magnetic force exerted on the moving charge by the current in the wire is radially inward
Explanation:
given information:
current, I = 3 A
= +6.5 x
C
r = 0.05 m
v = 280 m/s
and direction of the magnetic force exerted on the moving charge by the current in the wire, we can use the following formula:
F = qvB sin θ
where
F = magnetic force (N)
q = electric charge (C)
v = velocity (m/s)
θ = the angle between the velocity and magnetic field
to find B we use
B = μ
I/2πr
μ
= 4π x
or 1.26 x
N/
, thus
B = 4π x
x 3 / 2π(0.05)
= 1.2 x
T
Now, we can calculate the magnitude force
F = qvB sin θ
θ = 90°, because the speed and magnetic are perpendicular
F = 6.5 x
x 280 x 1.2 x
sin 90°
= 2.18 x
N
Using the hand law, the magnetic direction is radially inward