Physics: Post your doubts here!

[Saad the Paki]

Can someone tell me if this graph for a thermistor is correct?
Shouldn't it be like this?

 

[TariqBhai]

Can someone tell me if this graph for a thermistor is correct?View attachment 60441
Shouldn't it be like this?
View attachment 60442

You are correct.

 

[TariqBhai]

Part C
Rizwan Javed

 

[Rizwan Javed]

Part C
Rizwan Javed

Is this related to AS or A2?

 

[TariqBhai]

Is this related to AS or A2?

Circular Motion A2

 

[Rizwan Javed]

Circular Motion A2

Sorry, then I can't help since I'm doing AS
May be anyone else can help.

 

[TariqBhai]

Sorry, then I can't help since I'm doing AS
May be anyone else can help.

Can you tag someone you know?

 

[Rizwan Javed]

Can you tag someone you know?

Midnight dream My Name Physicist

 

[HumptyR]

R = V/I
I is 1.2 A
V is 12V
Find the total resistance from this which comes out to be 10 ohms
Now total resistance = resistance of X + resistance of Y + internal resistance
10 = 6 + Y +0.5
Y = 3.5 ohm
the resistance of X is found from the graph.

WHERE DID U GET THE TOTAL RESISTANCE FROM?

 

[Midnight dream]

Midnight dream My Name Physicist

I cant see the pic he just attached so pardon i cant help.

 

[Saad the Paki]

WHERE DID U GET THE TOTAL RESISTANCE FROM?

Total resistance = emf/current in circuit
R = 12/1.2 = 10 ohms

 

[ashcull14]

View attachment 60445

Part C

idk if im right but i solved it like this
Tcos0 = W
Tsin0 = Fc
Fc = 0.15*6.35^2/1.82 = 3.33N
make T the subject T= 1.47 / cos0
substitute it into Tsin0 = Fc
1.47/cos0 *sin0 = 3.33
tan0 = 3.33/1.47
0 = 66.0
Tcos66 = 1.47
T = 3.6..N

 

[Copy Cat]

Part C
Rizwan Javed

Can you help me with Part (a)?

 

[ashcull14]

Can you help me with Part (a)?

v = rw
w = 2πf
f = 10 / 18
= 2π(10/18) =3.49 m/s
v= 3.49 * 1.82 = 6.35 m /s

 

[funky brat]

Can you help me with Part (a)?

Find time period which is 18/10= 1.8 seconds
then use formula omega= 2*pie/T
2*pie/1.8
then use formula v=r*omega where r= 1.82m to get 6.35 as the answer

 

[funky brat]

How are the frictional forces related to the sharpness of the peak?

 

[Gambhir Bidari]

Could anyone please help me to figure out the solution for 9702 2009 May June Paper 21 Q no. 5, part 'b'?? I had problem understanding mark scheme so, I am for looking solutions. Please someone upload the solution understandably..

 

[TariqBhai]

View attachment 60445
View attachment 60457
idk if im right but i solved it like this
Tcos0 = W
Tsin0 = Fc
Fc = 0.15*6.35^2/1.82 = 3.33N
make T the subject T= 1.47 / cos0
substitute it into Tsin0 = Fc
1.47/cos0 *sin0 = 3.33
tan0 = 3.33/1.47
0 = 66.0
Tcos66 = 1.47
T = 3.6..N

I get the same answer 3.64 but the answer mentioned is 3.72? I think the mentioned answer is wrong.

 

[funky brat]

w15 paper 41 q 11ci, anyone?

 

[ashcull14]

I get the same answer 3.64 but the answer mentioned is 3.72? I think the mentioned answer is wrong.

might b bcz the angle of horizontal is correct...66