Lower Secondary Science

[pchong]

[1) Longman 1001 MCQs (http://www.pearson.sg/cos/o.x?c=%2Fpearson%2Flocallib&ptid=176&pg=1&sort=&subj=Physics+&level=&kw=mcq&type=title&ipp=20)


2) Smashing Exams (http://smashingexams.com.sg/)

The Marshall Cavendish Science guidebooks aren't too good though, in my opinion. I don't find the materials inside any different from the textbooks.[/quote]


Thank you so much for your help.
Will take a look at the 2 books that you recommend .[/quote][/quote]
Thanks for the sharing

[archie2]

physic Question


A block resting on a smooth horizontal floor is being pushed by a constant horizontal force.
the answer states that if the velocity at 1s is v, then the velocity at 2s will be 2v and the velocity at 3s will be 3v.
can someone explain why the velocity change from v to 2v and 3v ?
thank you
rgds

[Oldschool]

archie2:

physic Question


A block resting on a smooth horizontal floor is being pushed by a constant horizontal force.
the answer states that if the velocity at 1s is v, then the velocity at 2s will be 2v and the velocity at 3s will be 3v.
can someone explain why the velocity change from v to 2v and 3v ?
thank you
rgds

Hi,

This has to do with the understanding of Newton's 2nd law of motion, which states that,

\"The change in velocity (acceleration) with which an object moves is directly proportional to the magnitude of the force applied to the object and inversely proportional to the mass of the object.\"

Mathematically, Force = mass x acceleration

and Acceleration = (Final Velocity - Initial Velocity) / time interval

So for a constant force, the change in velocity or acceleration is also constant.

Therefore, resulted in the above statement \" if the velocity at 1s is v, then the velocity at 2s will be 2v and the velocity at 3s will be 3v\"

This can be illustrated as below,

http://i43.tinypic.com/312hj6d.jpg\">

Regards

[Vivian22]

Hi,


I have two Physics question to ask. Can someone please help? Thanks in advance!
http://i41.tinypic.com/ilf4wh.png\">
http://i41.tinypic.com/9ier77.png\">

[Dr.033430Daniel]

[quote=\"Vivian22\"]Hi,


I have two Physics question to ask. Can someone please help? Thanks in advance!
http://i41.tinypic.com/ilf4wh.png\">
For this first question, the result for lamp P is the clearest. I always think : What happens to the current and what happens to the voltage for each lamp. When lamp R is removed from the circuit the overall resistance of the circuit goes up. This is because the parallel combination of R and Q is lower resistance then just having Q in the circuit. So if the overall resistance of the circuit goes up, then the current goes down. So the current through P goes down. What about the voltage across P? It also goes down. The resistance of the rest of the circuit goes up, so the fraction of voltage dropped across P goes down. So P gets dimmer. Choice C is the only choice saying that.
For Q the voltage across it goes up when the parallel resistor R is removed. The current through Q therefore also goes up from V=IR. So bulb Q gets brighter. Answer: C.
I take usually 10 minutes to explain these types of problems in class, so I have to give the condensed version here:)

[Vivian22]

Dr.Daniel:

Vivian22:

Hi,


I have two Physics question to ask. Can someone please help? Thanks in advance!
http://i41.tinypic.com/ilf4wh.png\">
For this first question, the result for lamp P is the clearest. I always think : What happens to the current and what happens to the voltage for each lamp. When lamp R is removed from the circuit the overall resistance of the circuit goes up. This is because the parallel combination of R and Q is lower resistance then just having Q in the circuit. So if the overall resistance of the circuit goes up, then the current goes down. So the current through P goes down. What about the voltage across P? It also goes down. The resistance of the rest of the circuit goes up, so the fraction of voltage dropped across P goes down. So P gets dimmer. Choice C is the only choice saying that.
For Q the voltage across it goes up when the parallel resistor R is removed. The current through Q therefore also goes up from V=IR. So bulb Q gets brighter. Answer: C.
I take usually 10 minutes to explain these types of problems in class, so I have to give the condensed version here:)

:thankyou:

[dioprem]

Vivian22:

Hi,


I have two Physics question to ask. Can someone please help? Thanks in advance!
http://i41.tinypic.com/9ier77.png\">

I'll try to answer this question and minimise calculations as much as I can. (Not easy to type in Math in forums).

The bulbs are rated at 6V, 12W. In other words, it would require a potential difference of 6V across each bulb for the bulb to work normally. And when it does so, it dissipates 12W of power. With that information, we know that the bulb has a resistance of 3 ohms. (P=V^2/R). With this information, lets look at each option.

A) If the current though L1 is 2A, it would imply that the current through L2 is 2A as well. This would mean that a current of 4A is passing through L3. Using V=RI, it would imply that the potential difference across L3 is 12V, which is not possible since the e.m.f of the battery is only 6V. So this cannot be the answer.

B) If the potential difference across L2 is 3V, then using V=RI would show that the current through L2 is 1A. Similarly the current through L1 would also be 1A since it is in parallel. This would mean that the current through L3 is 2A. Using V=RI again, this would mean that the potential difference across L3 is 6V, which cannot be the case as the e.m.f of the battery is 6V, which would mean that the potential difference across L1 and L2 is 0V, which is not possile.

C) The effective resistance of the whole circuit is 4.5 ohms. Since the e.m.f of the battery is 6V, the current drawn from the battery is 1.33A. Hence by using P=IV, the power drawn from the battery is 8W. So this answer is also worng. (Also by doing this from the beginning, you would know from circuit analysis the the current drawn from the battery is the same as the current through L3. You could have used this method to eliminate A and B as the answers as well.

D)Using P=(I^2)R, the power dissipated by L3 is 5.33W. Similarly, the current drawn by L1 = current drawn by L2 = 1.33/2. Hence the power dissipated by L1 and L2 is less.

Hope this helps.

[Vivian22]

dioprem:

Vivian22:

Hi,


I have two Physics question to ask. Can someone please help? Thanks in advance!
http://i41.tinypic.com/9ier77.png\">

I'll try to answer this question and minimise calculations as much as I can. (Not easy to type in Math in forums).

The bulbs are rated at 6V, 12W. In other words, it would require a potential difference of 6V across each bulb for the bulb to work normally. And when it does so, it dissipates 12W of power. With that information, we know that the bulb has a resistance of 3 ohms. (P=V^2/R). With this information, lets look at each option.

A) If the current though L1 is 2A, it would imply that the current through L2 is 2A as well. This would mean that a current of 4A is passing through L3. Using V=RI, it would imply that the potential difference across L3 is 12V, which is not possible since the e.m.f of the battery is only 6V. So this cannot be the answer.

B) If the potential difference across L2 is 3V, then using V=RI would show that the current through L2 is 1A. Similarly the current through L1 would also be 1A since it is in parallel. This would mean that the current through L3 is 2A. Using V=RI again, this would mean that the potential difference across L3 is 6V, which cannot be the case as the e.m.f of the battery is 6V, which would mean that the potential difference across L1 and L2 is 0V, which is not possile.

C) The effective resistance of the whole circuit is 4.5 ohms. Since the e.m.f of the battery is 6V, the current drawn from the battery is 1.33A. Hence by using P=IV, the power drawn from the battery is 8W. So this answer is also worng. (Also by doing this from the beginning, you would know from circuit analysis the the current drawn from the battery is the same as the current through L3. You could have used this method to eliminate A and B as the answers as well.

D)Using P=(I^2)R, the power dissipated by L3 is 5.33W. Similarly, the current drawn by L1 = current drawn by L2 = 1.33/2. Hence the power dissipated by L1 and L2 is less.

Hope this helps.

That was a very clear explanation. Thank you! :goodpost:

[Crester]

Can anyone recommend Physics home tutor who can teach Upper Sec Pure Physics, please?

PM me.
TQ

[dioprem]

Vivian22:

dioprem:

[quote=\"Vivian22\"]Hi,


I have two Physics question to ask. Can someone please help? Thanks in advance!
http://i41.tinypic.com/9ier77.png\">

I'll try to answer this question and minimise calculations as much as I can. (Not easy to type in Math in forums).

The bulbs are rated at 6V, 12W. In other words, it would require a potential difference of 6V across each bulb for the bulb to work normally. And when it does so, it dissipates 12W of power. With that information, we know that the bulb has a resistance of 3 ohms. (P=V^2/R). With this information, lets look at each option.

A) If the current though L1 is 2A, it would imply that the current through L2 is 2A as well. This would mean that a current of 4A is passing through L3. Using V=RI, it would imply that the potential difference across L3 is 12V, which is not possible since the e.m.f of the battery is only 6V. So this cannot be the answer.

B) If the potential difference across L2 is 3V, then using V=RI would show that the current through L2 is 1A. Similarly the current through L1 would also be 1A since it is in parallel. This would mean that the current through L3 is 2A. Using V=RI again, this would mean that the potential difference across L3 is 6V, which cannot be the case as the e.m.f of the battery is 6V, which would mean that the potential difference across L1 and L2 is 0V, which is not possile.

C) The effective resistance of the whole circuit is 4.5 ohms. Since the e.m.f of the battery is 6V, the current drawn from the battery is 1.33A. Hence by using P=IV, the power drawn from the battery is 8W. So this answer is also worng. (Also by doing this from the beginning, you would know from circuit analysis the the current drawn from the battery is the same as the current through L3. You could have used this method to eliminate A and B as the answers as well.

D)Using P=(I^2)R, the power dissipated by L3 is 5.33W. Similarly, the current drawn by L1 = current drawn by L2 = 1.33/2. Hence the power dissipated by L1 and L2 is less.

Hope this helps.

That was a very clear explanation. Thank you! :goodpost: [/quote]No worries.