Physics: Post your doubts here!

[nehaoscar]

View attachment 54698
(ANS : D )

How ?

Use vertical motion to find the time (time to travel vertical distance = time to travel horizontal distance)
s = 1.25 , a = 9.81 , u = 0
using s = ut + 1/2 at^2
you get t = 0.5048...

Then use horizontal motion
s = 10 and t = 0.5048...
speed = distance / time (since there is no acceleration in horizontal motion and speed remains the same throughout)
this will give v = 19.8.... which is roughly 20.

 

[qwertypoiu]

I get it a bit, What I still don't get is why wud the pattern be like up down down up.. but it s okay thanks a lot !

Let me just make a last attempt
Since it's travelling at 4m/s, the wave must travel, in 0.125s, 0.5m. Let's look at the wave step by step: (s is how much wave moved horizontally from initial position)

t=1, (s=4)


x hasn't moved yet. It's displacement = 0

t=1.125, s=4.5

x has moved down. It's displacement = -1

T=1.25, s=5

At this moment, displacement is still -1, but within this instance, it's gonna go all the way up.

T=1.375, s=5.5

x has moved up, displacement = +1

T=1.5, s=6

x is at +1 but is gonna move down in this very instance.

Now you may plot time against displacement. You'll see that it matches graph of B

 

[awesomaholic101]

.
Why not B?

I guess the difference in angles between maximas should be same.

 

[Mahnoorfatima]

Please could you explain why the current is unchaged though as I don't understand that one? Like the total resistance of the circuit decreases, so the current increases. The ratio of the resistances at the junction would increase so more current would flow to the fixed resistor at the top so shouldn't ammeter reading increase?
But then why is it unchanged?

Look at the position of the ammeter. It's connected next to the fixed resistor and the fixed resistor is parallel to the variable one therefore the currents have to be different in both the loops i.e the ammeter is only giving you the current across the fixed resitance.

 

[ashcull14]

 

[nehaoscar]

I guess the difference in angles between maximas should be same.

Why does it have to? :/

 

[nehaoscar]

Look at the position of the ammeter. It's connected next to the fixed resistor and the fixed resistor is parallel to the variable one therefore the currents have to be different in both the loops i.e the ammeter is only giving you the current across the fixed resitance.

But then why doesn't the current through the fixed resistor change? Like the current is different in both loops
Now the resistance of the bottom loop is changed so it increases right
The top loop resistance stays the same
So the ratio of the 2 resistances in the 2 loops increases
The current divides in the inverse of the ratio of this resistance
So the current in the top loop would increase right? Like accross the fixed resistor... so wouldn't ammeter reading increase? :/

 

[Mahnoorfatima]

But then why doesn't the current through the fixed resistor change? Like the current is different in both loops
Now the resistance of the bottom loop is changed so it increases right
The top loop resistance stays the same
So the ratio of the 2 resistances in the 2 loops increases
The current divides in the inverse of the ratio of this resistance
So the current in the top loop would increase right? Like accross the fixed resistor... so wouldn't ammeter reading increase? :/

No! See you can't just change the resisitance of the fixed resistance if the current flowing through it is the same as it was before. Ratio of the resisitances does not increase in both of the loops. How can it? You shoud treat these loops seperately as seperate circuits. The main thing is the position of the ammeter, the resistance through it has to remain the same. If the ammeter was connected with both of the resisitors your theory would've been right but these are seperate two loops with seperate currents. Idk how to explain it. :/

 

[nehaoscar]

No! See you can't just change the resisitance of the fixed resistance if the current flowing through it is the same as it was before. Ratio of the resisitances does not increase in both of the loops. How can it? You shoud treat these loops seperately as seperate circuits. The main thing is the position of the ammeter, the resistance through it has to remain the same. If the ammeter was connected with both of the resisitors your theory would've been right but these are seperate two loops with seperate currents. Idk how to explain it. :/

Oh ok! So if the ammeter was positioned say between the two loops , then the reading would decrease?

 

[princess Anu]

Let me just make a last attempt
Since it's travelling at 4m/s, the wave must travel, in 0.125s, 0.5m. Let's look at the wave step by step: (s is how much wave moved horizontally from initial position)

t=1, (s=4)
View attachment 54723

x hasn't moved yet. It's displacement = 0

t=1.125, s=4.5
View attachment 54724
x has moved down. It's displacement = -1

T=1.25, s=5
View attachment 54725
At this moment, displacement is still -1, but within this instance, it's gonna go all the way up.

T=1.375, s=5.5
View attachment 54726
x has moved up, displacement = +1

T=1.5, s=6
View attachment 54727
x is at +1 but is gonna move down in this very instance.

Now you may plot time against displacement. You'll see that it matches graph of B

I got it !!
JazakAllah

 

[Turki AbdulAziz]

Q.P http://maxpapers.com/wp-content/uploads/2012/11/9702_w12_qp_13.pdf
M.S http://maxpapers.com/wp-content/uploads/2012/11/9702_w12_ms_13.pdf

Can someone please explain Q 25?

And for Q 38, I get how gamma radiation will not be stopped by a few millimeters of lead, but inst alpha radiation absorbed by a few centimtres of air (1cm-3cm) ?

 

[Ahmed Aqdam]

But in parallel the formula is 1/R = 1/R1 + 1/R2 etc... so if you do I/(1/R1 + 1/R2 ...) then it will give you IR right?

That would be if all resistors were in parallel but there is a resistor in series too.

 

[eliyeap]

Q.P http://maxpapers.com/wp-content/uploads/2012/11/9702_w12_qp_13.pdf
M.S http://maxpapers.com/wp-content/uploads/2012/11/9702_w12_ms_13.pdf

Can someone please explain Q 25?

And for Q 38, I get how gamma radiation will not be stopped by a few millimeters of lead, but inst alpha radiation absorbed by a few centimtres of air (1cm-3cm) ?

The weight of the man is supported by the 2 cables. Tension in 1 cable= (80*9.81)/2 = 392.4N
Stress= 392.4/ 1/4*3.142* 0.005^2
Strain= 0.001/10
E=2.0*10^11 Pa

 

[Sarosh Jameel]

Y

Is it C?

yes

 

[69ermalik007]

Help needed....

 

[69ermalik007]

 

[princess Anu]

answer?

View attachment 54764

 

[princess Anu]

A water cannon directs a jet of water towards a vertical wall. 300 kg of water hit the wall each minute. The water hits the wall horizontally with a velocity 20 m s–1. Assume the water falls vertically after hitting the wall.

What force does the water exert on the wall?A 100N B 200N C 3000N D 6000N
helpp!!!

 

[princess Anu]

Student uses a digital ammeter to measure a current. Reading of the ammeter is found to fluctuate between 1.98A and 2.02A. 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

Help please

 

[Xylferion]

Student uses a digital ammeter to measure a current. Reading of the ammeter is found to fluctuate between 1.98A and 2.02A. 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

Help please

The readings are between 1.98 and 2.02, however the important part in the question is that these values are to +- 1%.

So it's 1.98 +- 1% and 2.02 +- 1%.

Since the readings " fluctuate " betweem 1.98 and 2.02, find the average value. This is 1.98 + 2.02 / 2 = 2.00, as stated in the answers.

Now write out the percentage uncertainties as actual uncertainties.

1/100 * 1.98 = 0.02
1/100 * 2.02 = 0.02

When finding the average, we added the two values. So we do the same with the actual uncertainties, giving us 0.04.
This leads to 2.00 +- 0.04, so the answer is D.

Hope that helped!