Physics: Post your doubts here!

[nehaoscar]

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Can anyone please help and tell me why there is deviation in the value of centripetal force and gravitational force cause wasn't it that the gravitational force only provides the centripetal force?

I remember someone had posted this, I can't remember tho? Pleaseee help.
Ms
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I'm not sure if this is correct but maybe it's because the mass is on the equator i.e placed on the surface of the earth so centripetal force will be very small
cause cf = gf for a satellite above the earth's surface so in this case the mass is not above but on the surface if that seems logical?

 

[Catherine_1]

I'm not sure if this is correct but maybe it's because the mass is on the equator i.e placed on the surface of the earth so centripetal force will be very small
cause cf = gf for a satellite above the earth's surface so in this case the mass is not above but on the surface if that seems logical?

Yeah it kinda does since mv^2 is small and R is huge it is obvious it won't be equal. Thank youu.



In this the last part when the output is negative shouldn't diode B be conducting rather than G.:/

Ms


Thanks

 

[nehaoscar]

Yeah it kinda does since mv^2 is small and R is huge it is obvious it won't be equal. Thank youu.

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In this the last part when the output is negative shouldn't diode B be conducting rather than G.:/

Ms
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Thanks

Is this from the specimen paper??
And yes it is B

 

[Catherine_1]

Is this from the specimen paper??
And yes it is B

Yeah but the subsequent answer also mentions the same thing. :/ You're sure, no?



How does one ideally find out the mean pd from the graph tho?
I am not too sure.

 

[nehaoscar]

Yeah but the subsequent answer also mentions the same thing. :/ You're sure, no?

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How does one ideally find out the mean pd from the graph tho?
I am not too sure.
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yep... which website did you get the ms from cause the ms i have from the cie website says :

 

[nehaoscar]

Yeah but the subsequent answer also mentions the same thing. :/ You're sure, no?

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How does one ideally find out the mean pd from the graph tho?
I am not too sure.
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mean pd = 0.5(minpd + maxpd)
so basically just the average

 

[OmarKhan99]

Ah yeah, it's this question.

Once the string is cut, the log doesn't immediately fall down. (I believe you understood that.) In the log's motion upward after the string is cut, there is friction acting on the same side as the component of weight - along the slope. So the addition of both forces divided by the mass of the log would give you acceleration.
On the other hand, once the log reaches a zero velocity, that is the maximum height it can go up to. Then, it starts falling down. In this part of its motion, the component of weight along the slope is opposed by the friction. So, acceleration would be the difference in both forces over the mass of the log.
Hence, the two accelerations are different and you get two different v-t gradients.

Hope you got it.

Thanks once again. I know that acceleration of the log while traveling up an down the slope will be different. What I to meant to ask was, why are the accelerations different when traveling down the slope? In other words, why are the gradients of the red and yellow line different. Shouldn't it be a straight line?

 

[Catherine_1]

mean pd = 0.5(minpd + maxpd)
so basically just the average

Omg thank you so much.

yep... which website did you get the ms from cause the ms i have from the cie website says :
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Wait wut. Lol I got from papersgceguide site .
Lol thanks, I will check it out once again now.

 

[nehaoscar]

Omg thank you so much.

Wait wut. Lol I got from papersgceguide site .
Lol thanks, I will check it out once again now.

maybe it was some old copy which cie later corrected and republished i guess

 

[Catherine_1]

maybe it was some old copy which cie later corrected and republished i guess

Lol I guess

 

[Catherine_1]

How would the graph be?

 

[Catherine_1]

Would the answer to this question be correct in this way? Tho neater and clearer.

 

[Wkhan860]

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How would the graph be?

Normal 'curving' graph sloping upwards initially and then becoming constant.
As it says that it should be 1.5 times the value it was at t(1/2), then if the curve was at n=1 (first line of y-axis considered 1) at t(1/2) then at 2t(1/2) it should be at n=1.5 (middle of 1st nd 2nd line of y axis).

 

[Catherine_1]

Normal 'curving' graph sloping upwards initially and then becoming constant.
As it says that it should be 1.5 times the value it was at t(1/2), then if the curve was at n=1 (first line of y-axis considered 1) at t(1/2) then at 2t(1/2) it should be at n=1.5 (middle of 1st nd 2nd line of y axis).

Thank you so much. But actually I didn't understand why would it be 1.5 times, if you could elaborate please?

 

[Catherine_1]

Please explain.

 

[Konstantino Nikolas]

Thanks once again. I know that acceleration of the log while traveling up an down the slope will be different. What I to meant to ask was, why are the accelerations different when traveling down the slope? In other words, why are the gradients of the red and yellow line different. Shouldn't it be a straight line?
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red line does NOT depict log going down ... the log is still going up but with DECREASING velocity (negative gradient = deceleration) ... so until the red line gets over, the velocity is still positive as log is in upward motion. When log comes down, velocity is negative (yellow line). So now read that explanation again to understand why the magnitudes of acceleration of red and yellow lines are different.
You have misunderstood the motion of the log ... just focus and try to imagine the motion of the log ... read the explanation and you will hopefully get it.

 

[Catherine_1]

Please help, anyone?

 

[holoholo]

How do you answer this ?

 

[OmarKhan99]

red line does NOT depict log going down ... the log is still going up but with DECREASING velocity (negative gradient = deceleration) ... so until the red line gets over, the velocity is still positive as log is in upward motion. When log comes down, velocity is negative (yellow line). So now read that explanation again to understand why the magnitudes of acceleration of red and yellow lines are different.
You have misunderstood the motion of the log ... just focus and try to imagine the motion of the log ... read the explanation and you will hopefully get it.

Ohhhh!!! I GET IT!! Thanks a million bro!

 

[Konstantino Nikolas]

Ohhhh!!! I GET IT!! Thanks a million bro!

Good. No problem!