The document provides tips for taking online classes effectively. It recommends being prepared before class starts, treating the online course seriously, holding oneself accountable, practicing time management, creating a dedicated study space, eliminating distractions, figuring out how you learn best, actively participating, and leveraging your network. It also lists tips for inside the online class, such as opening your camera, muting audio when not speaking, taking notes on paper rather than the screen, asking permission before sharing, and being disciplined.

1. TIPS FOR TAKING ONLINE CLASS
FOR YOURSELF
1: Be ready before the starting time
2: Treat an online course like a real course
3: Hold yourself accountable
4: Practice time management
5: Create a regular study space and stay organized
6: Eliminate distractions
7: Figure out how you learn best
8: Actively participate
9: Leverage your network
INSIDE THE CLASS
1: Open your camera
2: Mute your audio
3: take your pen, copy and other material sources
3: Don’t add your note on the screen
4: Do ask, write and share after the permission. It
will be provided at the end of the class
5: Be more disciplined.
1

2. This is a REGULAR CLASS. Please make a
note by writing down what you have
seen here.
Make a separate note for CURRENT
ELECTRICITY.

3. Scalars & Vectors OLD is
GOLD Discussion

6. 𝐻𝑒𝑟𝑒 𝐴𝐵 ≠ 0 𝑡ℎ𝑒𝑛 𝑖𝑡 𝑚𝑢𝑠𝑡 𝑏𝑒
𝑐𝑜𝑠𝜃 = 0
→ 𝜃 = 90°

30. Numerical Problems

31. A truck travelling at a 50 km/h, north turns west at the same speed. What is the
change in velocity?
50 km/h
𝑽𝑨
𝑽𝑩
50 km/h
Change in velocity =∆𝑽 = 𝑽𝑩 − 𝑽𝑨
𝑽𝑩
𝑽𝑨
−𝑽𝑨
. . .. . . . . . . . . .
.
..
.
.
.
.
.
.
.
.
.
.
∆𝑉 = 𝑉𝐴
2 + 𝑉𝐵2
= 𝟓𝟎 𝟐 km/hr.
𝑽𝑨
𝑽𝑩
−𝑽𝑨
.
..
.
.
.
.
.
.
.
.
.
.
Alternative

32. Due East
Towards East
𝜃
North of East by 𝜃
E
N
S
W
North East
45°
45°
E
N
S
W
30°
N
E
S
W
E
N
S
W
N
𝜃
East of North by 𝜃
E
S
W E
N
S
W

33. A sail boat sails 2km east then 4km southwest, then an additional distance is unknown direction.
Its final position is 5 km directly east of starting point, Find the magnitude and direction of third
leg of journey.
Solution
Here
Rx = Ax + Bx + Cx
5 = 2 + (-4sin45°) + Cx
∴ 𝐶𝑥 = 5.83 𝑘𝑚
Ry = Ay + By + Cy
0 = O + (-4cos 45°) + Cy
∴ 𝐶𝑦 = 2.83 𝑘𝑚
Now
C = 𝒄𝒙
𝟐 + 𝑪𝒚𝟐
= 5.83 2 + 2.83 2
= 6.48 km
Direction : 𝐭𝐚𝐧
_𝟏 𝑪𝒚
𝑪𝒙
= 𝑇𝑎𝑛
_1 2.83
5.83
= 25.89° North of East
Alternatively
𝐴 = 2𝑖
𝐵 = −4𝑠𝑖𝑛45°𝑖 − 4𝑐𝑜𝑠45°𝑗 = −2.82𝑖 − 2.83𝑗
𝐶 = 𝑢𝑛𝑘𝑛𝑜𝑤𝑛
𝑅 = 5𝑖
𝑺𝒊𝒏𝒄𝒆: 𝑹 = 𝑨 + 𝑩 + 𝑪
5𝑖 = 2𝑖 − 2.83𝑖 − 2.83𝑗 + 𝐶
𝐶 = 5 − 2 + 2.83 𝑖 + 2.83𝑗
𝐶 = 5.83𝑖 + 2.83𝑗
𝑁𝑜𝑤
𝑀𝑎𝑔𝑛𝑖𝑡𝑢𝑑𝑒 ∶ 𝑪 = 𝟓. 𝟖𝟑 𝟐 + 𝟐. 𝟖𝟑 𝟐 = 6.48 𝑘𝑚
𝐷𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛: 𝑻𝒂𝒏
_𝟏 𝑪𝒚
𝑪𝒙
= 𝑇𝑎𝑛
_1 2.83
5.83
=25.89° North of East

34. Of west

38. Ans:

39. X
Y

41. UDAY KHANAL
Department of Physics
CCRC
41

42. The essence of SCIENCE: ask an impertinent question, and
you are on the way to a pertinent answer.

43. DID YOU ENJOYE THE CLASS?
Leave your valuable suggestions so that I will be
better for you all in the next class. Your
suggestions are highly appreciated.
NO?
Yes?