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Urgent PHY help needed..!

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hey plz anyone explain me the terminal velocity.. its out of mind.. plzz...
 
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I'll give you an example. When a ball is dropped from a certain height, its speed will increase. But over time, the upward force (air resistance) will have the same magnitude as the downward force (weight) so the forces are balanced and it moves at a constant speed. This constant speed = terminal velocity. The net force is zero, resulting in an acceleration of zero.

For that MCQ, when you drop the ball in air the velocity will increase so cross out option A and B. The initial acceleration of the ball is 10ms-2 but it'll soon decrease to zero. Since it's decreasing, option C is right.
 
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leosco1995 said:
I'll give you an example. When a ball is dropped from a certain height, its speed will increase. But over time, the upward force (air resistance) will have the same magnitude as the downward force (weight) so the forces are balanced and it moves at a constant speed. This constant speed = terminal velocity. The net force is zero, resulting in an acceleration of zero.

For that MCQ, when you drop the ball in air the velocity will increase so cross out option A and B. The initial acceleration of the ball is 10ms^-2 but it'll soon decrease to zero. Since it's decreasing, option C is right.


but y is velocity increasing and acceleration decreasing? :Search:
 
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The initial acceleration is 10ms-2. Eventually the acceleration will go to zero (I proved this already in my terminal velocity explanation above as the forces will become balanced and the net force will be zero, hence the acceleration will be zero), so it will decrease (from 10 to 0). As for the velocity, its potential energy will get converted to kinetic energy and thus the speed increases, I think.
 
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As the ball falls, two forces act on it: Its weight acting downwards and air resistance acting upwards. When the ball is dropped, its initial acceleration is 10 m/s2 (acceleration due to free fall), which would obviously cause its velocity to increase. As the velocity increases, the air resistance on the ball also increases. The resultant force which is found by = weight - air resistance is gradually decreasing as the air resistance is increasing (and weight is constant)! So the acceleration decreases and eventually becomes 0 m/s2 when air resistance = weight. At this point, the ball is said to fall with terminal velocity. It has zero acceleration and constant speed.
:)
 
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adnanj said:
abcde said:
You're welcome! :)

cAN u plz explain mcq 6 and 8 http://www.xtremepapers.me/CIE/Cambridg ... 9_qp_1.pdf
MCQ 6 : In order to keep the coin in the same position on the turnable, a resultant force must act on it towards the centre of the circle (centripetal force). So A is correct.
MCQ 8 : Read the stem of the question very carefully. It says that the bus is just about to topple over, meaning that at this instance, its centre of mass must act through its base (it is still stable!). This automatically rules out the possibility of A and B being correct. Using a bit of common sense, you'll know that C is correct, not D, because the centre of mass has to be in the region where more mass is concentrated. :)
 
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abcde said:
adnanj said:
abcde said:
You're welcome! :)

cAN u plz explain mcq 6 and 8 http://www.xtremepapers.me/CIE/Cambridg ... 9_qp_1.pdf
MCQ 6 : In order to keep the coin in the same position on the turnable, a resultant force must act on it towards the centre of the circle (centripetal force). So A is correct.
MCQ 8 : Read the stem of the question very carefully. It says that the bus is just about to topple over, meaning that at this instance, its centre of mass must act through its base (it is still stable!). This automatically rules out the possibility of A and B being correct. Using a bit of common sense, you'll know that C is correct, not D, because the centre of mass has to be in the region where more mass is concentrated. :)

y isn't B in mcq 8?? :unknown:
 
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If B is the centre of mass, the bus would topple over! Draw a vertical line from B downwards. You'll see that if B is the centre of mass, the weight of the bus will be acting outside the base of the bus, causing it to topple. However, the question mentions that the bus doesn't topple but is just ABOUT TO topple over. Got it?
 
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oh yeah got it... i didn't know about drawing a vertical line:) thank u for helping :D
 
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The answer is D. When both are joined together, their extension is x cm, so their separate extensions would be 2x cm. Since they are hanging one below the other, it'll be 2x + 2x = 4x cm.
 
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I think it's like this...? Anyhow, the angle of incidence should be equal to the angle of reflection so keep that in mind when making the actual ray.
 
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