Physics: Post your doubts here!

[sagar65265]

Please anyone explain me Q:14 . Thank a lot in advance
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_12.pdf

First point to note here - the block will only accelerate down the plane, not perpendicular to it. Therefore, there is only one component of acceleration, that is down the slope.

Let's apply Newton's Second Law down the inclined plane:

The component of gravity pulling the block down the plane is 10 Newtons.
The magnitude of friction resisting the downward motion of the block is equal to 4 Newtons.

Therefore, the net force is +10 + (-4) = 6 Newtons down the slope.

According to Newton's Second Law, we can write

6 Newtons = m * a

If the force of gravity on the object is 20 Newtons, and the acceleration due to gravity is 10 ms^-2, then we can write (using the relation Weight = mg)

20 Newtons = Mass * 10 ms^-2

Therefore, Mass = 2 kilograms.

So, 6 Newtons = 2 kilograms * a
Therefore, a = 3 ms^-2 = C.

Hope this helped!
Good Luck for all your exams!

 

[Thought blocker]

First point to note here - the block will only accelerate down the plane, not perpendicular to it. Therefore, there is only one component of acceleration, that is down the slope.

Let's apply Newton's Second Law down the inclined plane:

The component of gravity pulling the block down the plane is 10 Newtons.
The magnitude of friction resisting the downward motion of the block is equal to 4 Newtons.

Therefore, the net force is +10 + (-4) = 6 Newtons down the slope.

According to Newton's Second Law, we can write

6 Newtons = m * a

If the force of gravity on the object is 20 Newtons, and the acceleration due to gravity is 10 ms^-2, then we can write (using the relation Weight = mg)

20 Newtons = Mass * 10 ms^-2

Therefore, Mass = 2 kilograms.

So, 6 Newtons = 2 kilograms * a
Therefore, a = 3 ms^-2 = C.

Hope this helped!
Good Luck for all your exams!

Y u describe each and everything ?

 

[DeViL gURl B)]

31)
Total resistance in 6 and 3 ohm is 2 ohm, so the voltage will be 6V in d 2 ohm resistor. n 6V across d 6 n 3 ohm resistors respectively. we have the voltage n resistance. so V=I*R, 6=6*I. so current is 1A
View attachment 43614

So if the arrow was drawn in the 3 ohm thingi.. It'd be 6/3 ?
Btw thanks tons

 

[DeViL gURl B)]

q.4
% uncertainty = (uncertainty/reading) * 100
% uncertainty = (0.5/100-40)*100
% uncertainty = ± 0.83% <----- dis is da % uncertainty each time temperature is measured, so when it is measured twice (at 40 °c and 100°c) % uncertainty is added, so % uncertainty in % rise = 0.83% + 0.83% = 1.66% ≈ 1.7%
Answer: D

q.3
p = mv
p² = m²v²
p²/m = m²v²/m = mv² , dis is similar to da formula for KE except for da (1/2) , but becuz (1/2) is a constant n doesnt have any units, da base units of mv² n (1/2)mv² will b same, so answer is A: Energy

Thanks bro

 

[Thought blocker]

So if the arrow was drawn in the 3 ohm thingi.. It'd be 6/3 ?
Btw thanks tons

Of course ._.

 

[Hadi Murtaza]

Thanks bro

No broblem

 

[DeViL gURl B)]

papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Physics%20(9702)/9702_w10_qp_11.pdf

Question 5
Q.7 .. Doesn't the horizontal comp. stay constant through out ? :$
Q.9
Q.10 -_-
Q.11 .. Why isn't it A?
Please help !
Thank you

 

[DeViL gURl B)]

Of course ._.

It was a doubt .. Okha

 

[Thought blocker]

It was a doubt .. Okha

npnp ._.

 

[Thought blocker]

papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Physics%20(9702)/9702_w10_qp_11.pdf

Question 5
Q.7 .. Doesn't the horizontal comp. stay constant through out ? :$
Q.9
Please help !
Thank you

5)

---------->Written form :
Frequency: 3000 oscillations in a minute so , 3000/60sec= 50
now Time period= 1/50= 0.02
and so for time base, as this is for one full wave, divide it by 2, 0.02/2= 0.01

7)The horizontal velocity will decrease to 0 as air resistance is acting against it so it will eventually decrease to 0 and the vertical velocity will increase until it reaches a constant value the question is actually related to terminal velocity theory

9)Use v^2 - u^2 = 2as
v has to be zero, since the train has to come to rest eventually.
s = x
The equation becomes: 2ax + u^2 = 0
make 's' distance the subject:
x = - (u^2)/2a
deceleration, i.e -a, is constant in the question.
take the constants out,
you get this relation: x ∝ u^2, i.e, s varies as the square of u.
So when u is increased by 20% (120/100 = 1.2), x will change by (1.20)^2 = 1.44x

 

[sagar65265]

Y u describe each and everything ?

Because this question uses basic concepts. It is very important to fully understand all the basic concepts, because the most complex problems can still use basic concepts. Maybe I overdid it on this one, but still - I read somewhere that PhDs teaching at Uni level teach freshers and first year students for the same reason - even though they are experts in the field, they teach the most basic courses because basic concepts are extremely important to get right. Everything else you learn is based on those, so if you understand them perfectly, your further studies become that more easier and better.

 

[Thought blocker]

Because this question uses basic concepts. It is very important to fully understand all the basic concepts, because the most complex problems can still use basic concepts. Maybe I overdid it on this one, but still - I read somewhere that PhDs teaching at Uni level teach freshers and first year students for the same reason - even though they are experts in the field, they teach the most basic courses because basic concepts are extremely important to get right. Everything else you learn is based on those, so if you understand them perfectly, your further studies become that more easier and better.

I want something from you, Check ua convo bro

 

[DeViL gURl B)]

papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Physics%20(9702)/9702_w10_qp_11.pdf

Question 5
Q.7 .. Doesn't the horizontal comp. stay constant through out ? :$
Q.9
Q.10 -_-
Q.11 .. Why isn't it A?
Please help !
Thank you

sagar65265 could u please help in this

 

[Thought blocker]

You din't got me ?

sagar65265 could u please help in this

 

[DeViL gURl B)]

You din't got me ?

No I got 5
But the rest of 'em u didn't reply ..so u didn't know right.. For the rest
That's why

 

[Thought blocker]

-_- Check it again ._. huh!

 

[DeViL gURl B)]

-_- Check it again ._. huh!

Oooohhh thaaakkkk youuuu I dint read the written thingi! Thank you loadz
Plus question 22 as well please

 

[Thought blocker]

22)The force acting on the metal X is the same as the force acting on the metal Y(denote it as F) and according to Hooke's law: lambda(modulus of elasticity)/natural length = F/x.
for metal X: lambda = 3*F/x = 3 m*F/1.5 mm= 2000F
for metal Y: lambda = 1*F/x = 1m* F/1.0 mm= 1000F

hence the extension for the second case is:

for metal X: x = F*natural length/(lambda) = F*1 / 2000 F = 0.5 mm
for metal Y: x = F*3 / 1000 F = 3 mm

The total extension is 3.5 mm

 

[DeViL gURl B)]

Thought blocker
Dude when they say twice the length and twice the radius..what'd be the resistance of the wire?
Won't it be R?

​

 

[Thought blocker]

Which question ._. ? ?