Physics: Post your doubts here!

[ashcull14]

Can u plz plz solve this question ... I would be grateful

is it C?

 

[Rockstar RK]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w02_qp_1.pdf
Q36

is the answer a?

 

[Hijab]

is it C?

Yup

 

[ashcull14]

if its C then lambda/2=15*10^-3
so wvelenght=0.03m
f=3*10^8/0.03=1*10^10hz

 

[ZaqZainab]

Please solve View attachment 45275

The answer is B.[/quote]
It says the sectional area
If you read the value on the micrometer it is 3.81
so 3.81*0.5 is the radius
they want the crossectional area pie r^2
pie*(3.81*0.5)^2

 

[ashcull14]

yes

is the answer a?

 

[Rockstar RK]

yes

See its lengthy but easy...
All u shld know is that when the connection is in parallel, voltage division is the same for both..(here 2V in both)
Also resistors are equally distributed in the loops

 

[Dr.MMM]

A student uses a digital ammeter to measure a current. The reading of the ammeter is found to
fluctuate between 1.98 A and 2.02 A.

The manufacturer of the ammeter states that any reading has a systematic uncertainty of ± 1 %.

Which value of current should be quoted by the student?
A (2.00 ± 0.01) A
B (2.00 ± 0.02) A
C (2.00 ± 0.03) A
D (2.00 ± 0.04) A

How is the answer D?

 

[Laila39]

Anyone help me with Q.14, 32, 33, 35. Please.

 

[ashcull14]

can someone solve it ive forgotten hw i did it bfore

 

[Laila39]

2.02-1.98 = 0.04

 

[Rockstar RK]

yes

let the left loop be called as A and the right loop as B(loop of the resistors)
For loop A, total of resist. is 15ohm and in other loop i.e. B is also 15ohm
But both of them are in parallel that why resistance of the circit as a whole is 15/2 (1/15+1/15=2/15 and its inverse)

 

[Wolfgangs]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w02_qp_1.pdf
q36 please

 

[Hijab]

if its C then lambda/2=15*10^-3
so wvelenght=0.03m
f=3*10^8/0.03=1*10^10hz

Why did u divide lambda by 2???

 

[MaboroshI_I]

Can I get help with this question please?

 

[Rockstar RK]

yes

using V=IR
2=I(15/2)
=> I=4/15
Dividing current equally, current in A is 2/15 and in B is 2/15
Now in a X is attached to end of 1 resistor, hence in A again using V=IR we getV=2/15*5=2/3
Now subtracting 2-2/3 gives 4/3 is the voltage at X
now, Y is attached to end of second resistor in loop B, hence, V at Y=2/15*10=4/3
Subtracting 4/3 from 2 gives 2/3
Now V at X - Vat Y= 4/3-2/3 =2/3

 

[danial 234]

View attachment 45277
can someone solve it ive forgotten hw i did it bfore

120-80=40 kg
force=40*9.81
f=ma
find acc .. then use equation of motion to find speed when d=9metres

 

[danial 234]

Can I get help with this question please?

as R and 2R are in parallel .. so V2=V3
V=V1+V2 or V=V1+V3
V-V1=V3

 

[Rockstar RK]

Why did u divide lambda by 2???

It is because of the formula that distance between two adjacent maxima is lambda/2 and distance between minima and maxima is lambda/4

 

[Thought blocker]

I am glad, we all are capable of getting at least 35 of 40