Physics: Post your doubts here!

[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

 

[Thought blocker]

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

Oh, I got A

 

[agha saad 22]

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

D*S*A kahan say derive kiya hai?

 

[usama321]

D*S*A kahan say derive kiya hai?

Check the units

 

[meerul264]

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

View attachment 39686anothe

I see, so you treat AC as if it's resistors combined in parallel?

And voltage dissipated across CB first? despite the direction conventional current?

 

[usama321]

I see, so you treat AC as if it's resistors combined in parallel?

And voltage dissipated across CB first? despite the direction conventional current?

Not dissipated first. But not all is equally distributed across the parallel wires. Some is left and dissipated across CB

 

[agha saad 22]

Check the units

naee i mean kin formulas say derive kiya hai?? like KAI VALUES KO DEKH KAI KHUD m/s nikala hai!!?

 

[Suchal Riaz]

does anyone want readings for practicals so he can practice out the working and graph area?

 

[Suchal Riaz]

Oh, I got A

you should be happy if you get A

 

[asma tareen]

does anyone want readings for practicals so he can practice out the working and graph area?

Meeeee ._.

 

[Suchal Riaz]

Meeeee ._.

okay dadi. it's uploading

 

[Thought blocker]

you should be happy if you get A

I mean, I got wrong answer. :'(

 

[Thought blocker]

okay dadi. it's uploading

Me too

 

[agha saad 22]

okay dadi. it's uploading

chemistry kai :3

 

[Suchal Riaz]

summer 2013 practicals 33
readings only
https://www.dropbox.com/s/sokx40nb7zdzuc8/S13-33.pdf