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Physics: Post your doubts here!

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someone please explain me these two questions:
  • A sprinter runs a 100 m race in a straight line. He accelerates from the starting block at a constant acceleration of 2.5 m s–2 to reach his maximum speed of 10 m s–1. He maintains this speed until he crosses the finish line. Which time does it take the sprinter to run the race? A 4s B 10s C 12s D 20s. Answer is C, how?
  • An insect jumps with an initial vertical velocity of 1.0ms–1, reaching a maximum height of 3.5 × 10–2 m. Assume the deceleration is uniform. What is the magnitude of the deceleration? A 3.6ms–2 B 9.8ms–2 C 14ms–2 D 29ms–2 . Answer is C, how?
Thank you c:
 
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someone please explain me these two questions:
  • A sprinter runs a 100 m race in a straight line. He accelerates from the starting block at a constant acceleration of 2.5 m s–2 to reach his maximum speed of 10 m s–1. He maintains this speed until he crosses the finish line. Which time does it take the sprinter to run the race? A 4s B 10s C 12s D 20s. Answer is C, how?
  • An insect jumps with an initial vertical velocity of 1.0ms–1, reaching a maximum height of 3.5 × 10–2 m. Assume the deceleration is uniform. What is the magnitude of the deceleration? A 3.6ms–2 B 9.8ms–2 C 14ms–2 D 29ms–2 . Answer is C, how?
Thank you c:
For second q. use v^2=u^2 + 2as
 
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Tysmm for the answerss and Can you explain a bit about how you knew that it will occur lamda/4 and lamda later ?

you want to find out when you will get the same length of amplitude as y1, so in this case Y1 MUST BE POSITIVE, and the same amplitude will thus occur one wavelength later as two points on a single transverse wave with the same phase difference will have the SAME AMPLITUDE (length from the origin line) (look at the second line in my diagram). The first line is 1/4 lambda later as two points 1/4 lamda later will also have the SAME AMPLITUDE (length from the origin line) .

I'm sorry but explaining this is a bit difficult for me, but I hope you get the idea by looking at my drawing. I'm sorry if I couldn't explain it to you.

I actually have to show you the diagram myself and talk to make you understand, maybe a video, but that's not possible! Words can only take you to till a limit!!

hope it helps you :)
 
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I
you want to find out when you will get the same length of amplitude as y1, so in this case Y1 MUST BE POSITIVE, and the same amplitude will thus occur one wavelength later as two points on a single transverse wave with the same phase difference will have the SAME AMPLITUDE (length from the origin line) (look at the second line in my diagram). The first line is 1/4 lambda later as two points 1/4 lamda later will also have the SAME AMPLITUDE (length from the origin line) .

I'm sorry but explaining this is a bit difficult for me, but I hope you get the idea by looking at my drawing. I'm sorry if I couldn't explain it to you.

I actually have to show you the diagram myself and talk to make you understand, maybe a video, but that's not possible! Words can only take you to till a limit!!

hope it helps you :)
I actually do get it now and I realize its so simple . Its just that as we move in the y direction at that very same displacement ....from the rest position (x line). Thankyouu!
 
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help me with question 325 please....how is upthrust greater than force P ? if upthrust is greater than dowmward force wouldn't the sphere float ? i went for option B ...answer is A
 

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help me with question 325 please....how is upthrust greater than force P ? if upthrust is greater than dowmward force wouldn't the sphere float ? i went for option B ...answer is A
The sphere will not move in a diagonal direction so S= Q
For an object submerged in water there will be resultant upthrust force acting on it causing it to move upwards so R>P IT SHOULD FLOAT (e.g place a tennis ball under water it floats.)
(B implies that sphere would sink which is not the case)
 
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http://papers.gceguide.com/A Levels/Physics (9702)/9702_m16_qp_12.pdf
Q37 anyone and 34..35

Beginning from Q 34:
this is a simple question which requires correct manipulation of kirchoff's 2nd law ;
e.m.f's on one side and the p.d. 's on another side
so if we move from V1 to V2 and then V3 we get sth like
V1 - V2 - V3 ...
and similarly on the other side we get potential drops as ;
IR1 + IR2 + IR3 CAN BE WRITTEN AS I (R1+ R2+ R3)
so consequently we get
V1 - V2 - V3 = IR1 + IR2 + IR3
or as
V1 - V2 - V3 = I (R1+ R2+ R3)

I think i made some argue ment
:)

 
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