Physics: Post your doubts here!

[TheManiac]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_qp_21.pdf
'Q5(iii)', why 1.03V?????

[URL='http://www.regentsprep.org/Regents/physics/phys03/bkirchof2/']2. Kirchhoff's Law[/URL]http://www.regentsprep.org/Regents/physics/phys03/bkirchof2/

 

[biscuitbiscuit]

2. Kirchhoff's Law

Yeahhh!!, thanks

 

[usama321]

thnx

but how do u know that the skid marks appeared when the breaks were applied

The questions says "
With the brakes applied, the front
wheels of the car leave skid marks on the road that are 12.8 m long, as illustrated in Fig. 2.1"

 

[Umer Jawed]

Assalamualaikum I have problem in J2011 P23 Q6b and c(iv). Please help me with the sketches. I would be glad!

 

[usama321]

help!

max time =
0 = 16 -9.81*t
t = 1.631s
t/2 = .815s
speed = 8 m/s
KE = 32m
s = ut -1/2gt^2
s =9.78m
Gp energy = 96m
96/32 = 3

 

[Thought blocker]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w13_qp_23.pdf
nov 2013 p23

For the ball to reach max height
0 = 16 - gt
t= 1.63s
So, t/2 = .815s
Now, speed at .815s
v = 16 -g*.815
v= 8m/s
Thus K.E at this point would be
32m. m= mass of object
Now calcualte height for mgh at this point and you will get
s = 16*.815 - 1/2 g *.815 square
s = 9.78
Calculate mgh
96m
96m/32m = 3:1 ty to usama321

 

[meerul264]

Here

thanks but i'm still confused a bit. can you please explain to me the full concept of this question (about potential divider?) ?

 

[amal sharkawi]

A strong wind of speed 33 m s–1 blows against a wall. The density of the air is 1.2 kg m–3. The wall
has an area of 12 m2
at right angles to the wind velocity. The air has its speed reduced to zero
when it hits the wall.

What is the approximate force exerted by the air on the wall?
A 330 N B 400 N C 480 N D 16 000 N

can someone explain me this

 

[asma tareen]

A strong wind of speed 33 m s–1 blows against a wall. The density of the air is 1.2 kg m–3. The wall
has an area of 12 m2
at right angles to the wind velocity. The air has its speed reduced to zero
when it hits the wall.

What is the approximate force exerted by the air on the wall?
A 330 N B 400 N C 480 N D 16 000 N

can someone explain me this

agha saad 22

 

[usama321]

thanks but i'm still confused a bit. can you please explain to me the full concept of this question (about potential divider?) ?

As CB also has resistance, just think of it as another resister in series with the two parallel resistors. now if the emf of the batter is 6v (dont remember the question), some of the voltage would be dissipated across the CB resistor first, and the rest of 6v is then equally dissipated across the two parallel resistors

anothe

 

[agha saad 22]

agha saad 22

SORRY FOR LATE ANSWER ASMA!! WO JANA PARA LET MEE SEE WATS IT!! MMM ANSWER "A" hai?

 

[asma tareen]

SORRY FOR LATE ANSWER ASMA!! WO JANA PARA LET MEE SEE WATS IT!! MMM ANSWER "A" hai?

I dunno the answer I think Its A thts y I asked u :/

 

[usama321]

A strong wind of speed 33 m s–1 blows against a wall. The density of the air is 1.2 kg m–3. The wall
has an area of 12 m2
at right angles to the wind velocity. The air has its speed reduced to zero
when it hits the wall.

What is the approximate force exerted by the air on the wall?
A 330 N B 400 N C 480 N D 16 000 N

can someone explain me this

Mass of wind coming in contact with the blades per second = density * speed * area = 475.2 kg/s

F = mv-mu/t As it is 475.2 kg/s, we will use
F = 475.2*0 - 475.2*33/1 = D

 

[Thought blocker]

Mass of wind coming in contact with the blades per second = density * speed * area = 475.2 kg/s

F = mv-mu/t As it is 475.2 kg/s, we will use
F = 475.2*0 - 475.2*33/1 = D

I too got D

 

[asma tareen]

Please explain

Mass of wind coming in contact with the blades per second = density * speed * area = 475.2 kg/s

F = mv-mu/t As it is 475.2 kg/s, we will use
F = 475.2*0 - 475.2*33/1 = D

 

[usama321]

Please explain

The above formula is for rate of flow. Using that, we know that 475.2kg of wind is coming in contact with the wind.

Now we also now that force = rate of change of momentum. The final velocity is zero thus the final momentum is zero. The initial momentum would be 475.2 * 33. From the rate of flow, we know that every second 475.2 kg of mass comes in contact with the blades and stops. Thus use rate of change of momentum with time being 1s due to the rate of flow

 

[asma tareen]

The above formula is for rate of flow. Using that, we know that 475.2kg of wind is coming in contact with the wind.

Now we also now that force = rate of change of momentum. The final velocity is zero thus the final momentum is zero. The initial momentum would be 475.2 * 33. From the rate of flow, we know that every second 475.2 kg of mass comes in contact with the blades and stops. Thus use rate of change of momentum with time being 1s due to the rate of flow

Got It thanks

 

[amal sharkawi]

A solid rubber ball has a diameter of 8.0 cm. It is released from rest with the top of the ball 80 cm
above a horizontal surface. It falls vertically and then bounces back up so that the maximum
height reached by the top of the ball is 45 cm, as shown.

80
60
40
20

If the kinetic energy of the ball is 0.75 J just before it strikes the surface, what is its kinetic energy
just after it leaves the surface?
A 0.36 J B 0.39 J C 0.40 J D 0.42 J

QUESTION 17 PAPER 11 JUNE13
ANY ONE CAN ANSWER THIS QUESTION

 

[usama321]

A solid rubber ball has a diameter of 8.0 cm. It is released from rest with the top of the ball 80 cm
above a horizontal surface. It falls vertically and then bounces back up so that the maximum
height reached by the top of the ball is 45 cm, as shown.

80
60
40
20

If the kinetic energy of the ball is 0.75 J just before it strikes the surface, what is its kinetic energy
just after it leaves the surface?
A 0.36 J B 0.39 J C 0.40 J D 0.42 J

QUESTION 17 PAPER 11 JUNE13
ANY ONE CAN ANSWER THIS QUESTION

The trick in this question is to remember that the height in mgh is from the bottom of the object. Thus, when using mgh, we will use 72 and 37cm as its height, subtracting the diameter. Thus
mgh = KE
9.81*.72*m = 0.75
m = .106g
.106* 9.81*.37 = KE
KE = .385 = B

 

[chocolate_girl]

-->Intensity is directly propotional to square of a
it means if A in two times, I is four times
-->Intensity is inversely proportional to area
by first rule and second rule, E increase 4 times due to doubling of A but decreases 1/2 times due to halving of area. so E(new) = E*4/2=2E

Thanks