How was M1 (Math P/42)? (No discussion)

[Shadow]

I'm afraid it will cost you marks then.. But you will get marks in the first part

 

[sara kamal]

I'm afraid it will cost you marks then.. But you will get marks in the first part

hmm..thanku!

 

[Shadow]

hmm..thanku!

No prob Hope for the best!

 

[sara kamal]

No prob Hope for the best!

Yea In sha Allah khair!

 

[Shadow]

Yea In sha Allah khair!

Inshaa Allah!

 

[hacker]

yr Q6 last part ki working samjha do kesa aya answer ?
Mass 0.2 kg say ziada ana chahiye :C

 

[qwertypoiu]

yr Q6 last part ki working samjha do kesa aya answer ?
Mass 0.2 kg say ziada ana chahiye :C

you mean q7??

 

[qwertypoiu]

yr Q6 last part ki working samjha do kesa aya answer ?
Mass 0.2 kg say ziada ana chahiye :C

FOR THE LAST PART




W = 2N as ring was 0.2kg.
Theta = tan^-1(30/40) = 36.87 degrees.
Resolving horizontal:
R = 3.36 x sin(theta) = 3.36 x 3/5 = 2.016N

Find friction:
Fr = meu x R = 0.343 x 2.016N = 0.6915N

Resolve vertical:
Fr + 3.36 x cos(theta) = W + mg
0.6915N + 3.36*(4/5) = 2N + 10m
3.3795N = 2N + 10m
1.3795N = 10m
m = 0.138kg

 

[hacker]

FOR THE LAST PART


View attachment 53651

W = 2N as ring was 0.2kg.
Theta = tan^-1(30/40) = 36.87 degrees.
Resolving horizontal:
R = 3.36 x sin(theta) = 3.36 x 3/5 = 2.016N

Find friction:
Fr = meu x R = 0.343 x 2.016N = 0.6915N

Resolve vertical:
Fr + 3.36 x cos(theta) = W + mg
0.6915N + 3.36*(4/5) = 2N + 10m
3.3795N = 2N + 10m
1.3795N = 10m
m = 0.138kg

thank u yr
i misread and thought that another ring m is attached and ignored 2N :/