Physics: Post your doubts here!

[Sarosh Jameel]

Hi, does anyone know how to solve this? How do you know which forces are going to the clockwise and the anti clockwise direction?? Please help Thank you!!!View attachment 51679

Clockwise moment = 300 * 0.8* = 240 Nm

Anticlockwise moment = 200*0.8 = 160 Nm

We need more anticlockwise moment to attain equilibrium..
THUS >> 240-160= 80 Nm

80 Nm anticlockwise will bring equilibrium to the system ...
THUS D !!

 

[princess Anu]

help!

 

[princess Anu]

Hi, does anyone know how to solve this? How do you know which forces are going to the clockwise and the anti clockwise direction?? Please help Thank you!!!View attachment 51679

What is the answer?

 

[manya]

can someone help me with this question

 

[Itsmeeee]

Can Someone Please Help Me?

 

[manya]

check q 41 at
http://physics-ref.blogspot.com/2014/11/physics-9702-doubts-help-page-7.html

thankyou so much but i dont understand becuse x is moving up and y is moving down. what would be the phase difference if the wave was like this

 

[The Sarcastic Retard]

Hi, does anyone know how to solve this? How do you know which forces are going to the clockwise and the anti clockwise direction?? Please help Thank you!!!View attachment 51679

Sum of clockwise moments = sum of anticlockwise moments.
0.4*300 = 0.8*200
120 = 160
Hence total torque produced will be 160 - 120 = 40 Anticlockwise = B

 

[The Sarcastic Retard]

help!

for q11 : Is answer D?
For q8 : Is it B?
For Q14 : C?

 

[The Sarcastic Retard]

Can Someone Please Help Me?

Increase in force = resultant force of y - of x
2[120cos(55)] – 2[100(cos(65)] = 53N

 

[manya]

it would still be zero. remember it's a stationary wave

oh okay thankyou so much

 

[manya]

thankyou so much but i dont understand becuse x is moving up and y is moving down. what would be the phase difference if the wave was like this

also can u help me with same question part c4 i understand the explanation u hv given but incase it was 0.5t will we draw the wave like this

 

[Physux]

Need help.

 

[princess Anu]

for q11 : Is answer D?
For q8 : Is it B?
For Q14 : C?

Q8 is A and Q11 is B while Q14 is C
Can u show how you solved Q14

 

[Itsmeeee]

Increase in force = resultant force of y - of x
2[120cos(55)] – 2[100(cos(65)] = 53N

Thank you soo much

 

[Xylferion]

help!

For Q8, regarding the ball. From X to Y, the ball accelerates uniformly till it reaches point Y, you could say that the weight of the ball is what's making it accelerate. When it passes Y, the ball is now on a horizontal surface, since weight acts vertically, the ball wont accelerate horizontally, so the graph is a straight line.

Now the important thing here is when the ball collides with the wall. Since it collides inelastically, kinetic energy is not conserved completely, this will affect the speed and make the ball travel slowly. They mention in the question, that after the collision the ball comes to rest momentarily somewhere on XY. Since the ball has to reach XY again, and this time with a lower speed, it's obviously going to take longer. Graphs B and D both indicate that it takes the same amount of time, which is not true, since to cover the same distance at a slower speed, will take a lot more time. This leaves us with C and A.

The difference in the two graphs is the steepness of the ends of the graphs. A steeper gradient means the acceleration is greater. If the ball is going to be approaching XY and come to rest, how can the acceleration to roll up be greater than when it was rolling down? This immediately eliminates C, and your answer is A.

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For Q14,
-You're told that the mass of the ruler is 100g, this will act at the halfway point of the ruler since it is a uniform ruler. So 100g acts at 50cm. Your pivot is at the 40cm mark.
That means that 100g acts downwards, 10 cm from the pivot.
-Next you're told that a mass of 20g acts around the pulley. This force is acting up since the string is pulling ruler up, as the weight of the mass acts down. This acts 60cm from the pivot, and acts upwards.
- The next and probably most important part of the question is them asking you where to " suspend " a mass of 50g to " balance " the rule. If the mass is to be suspended from the ruler it is to act downwards. Now the question is do we put it to the right or left of the pivot. To figure that out, we first check what the resultant moment is and in which direction.

That is ( 60 cm * 20g ) - ( 10 cm * 100g ) = 11760 - 9800 = 1960 Nm upwards.

Since the resultant moment is upwards, that means suspending the weight to the left of the pivot would be pointless. This would just make the ruler turn even more in the anti-clockwise direction. So this alone lets us know that the weight is to be placed to the right of the pivot, if we are to balance it.

Now you form your equations, keeping in mind that for the ruler to be in equilibrium, the resultant moment must equal zero.

Moments about Pivot: (100g x 10cm) + ( 50g x ycm ) - (20g x 60cm) = 0
This gives us: 9800 + 490y - 11760 = 0
490y = 1960 Ncm
y = 1960/490 = 4 cm

We now have our distance from the pivot, and since we know that the weight is to be suspended to the right, we add this distance to the position of the pivot.
40 cm + 4 cm = 44 cm, and thus your answer is C.

Hope that helps.

 

[manya]

yeah. it also said this in the explanation there

I read that, just wanted to confirm, thankyou so much

 

[Jennifer4678]

Thanks guys!!!!!

have you checked at
http://physics-ref.blogspot.com/2014/06/9702-november-2013-paper-41-42-worked.html

try to ask through the comment box there

 

[Physux]

Need help.

I need help people?!

 

[DeViL gURl B)]

http://maxpapers.com/syllabus-mater...hment/9702_may-june-2011-all-question-papers/

physics, May June 2011 paper 41 question 4 part (i) (ii) along with (c) .. can someone please EXPLAIN THIS QUESTION. ITS REALLY CONFUSING.
and in part (c) they've used the elementary charge but taken "-19" as "19" only .. in fact , all the negative values are taken as positive .. i don't get why.
CAN SOMEONE PLEASE HELP.
THANK YOU.

 

[Xylferion]

Need help.

Alright.. so for question 2, you need to find the torque. Torque is force x perpendicular distance between two forces. In that diagram you're already given the distance which is 0.30 m. However, the forces are not perpendicular, So you're going to want to find the vertical components. That is 2 Sin 50.

Now put that stuff into the formula, 2 Sin 50 x 0.30 m = 0.459 Nm. Which rounds up to 0.46 Nm.

---------------
I already explained question 6 above, have a read through it.
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For question 9,



Look at the diagram above, what I did was add the tails of each force, to the heads of the other.
You end up with a rhombus. Now the original two forces, the 10 N ones acting 120 degrees apart, act " outwards ". The resultant force will act outwards as well. Hence why I labelled that line running through the middle, R.

Now further inspection of R, should show you that it bisects 120* in half. This leaves 60 degrees on either side. Now that you have two equal angles and two equal sides, this should become more obvious to you :-

With this in mind, the final side will be of magnitude 10N and its counter-part angle will be 60 degrees.

NOTE:- You can tell it bisects in half if you touch up on the laws of a rhombus.

This isn't the only way to approach this question, but it's in my opinion, the easier approach. You could attempt to use a variation of the cosine rule, in which there is a slight difference in the sign usage.

Hope that helped. ^_^