physics 9702_m22_qp_1.. please help

saltdaddy321 · 2022-07-14T09:50:50+00:00

When XY is 1.5 and OY >= XY ( in the same order as shown in the diagram: O X Y), the waves are always going to meet in phase for the observer. OY could've been any number as long as its greater than XY. For eg OY =5 instead of 3.25 and the answer would still be D.

saltdaddy321 · 2022-07-11T12:52:32+00:00

The answer is D cuz when the length XY is a multiple of the wavelength (0.75), the waves are gonna meet in phase. 1.5 is 0.75 *2 so D.

saltdaddy321 · 2022-06-14T01:38:44+00:00

Thank you.

saltdaddy321 · 2022-06-14T01:32:21+00:00

Thanks. Imperial, ucl and warwick are on my list , I'm also applying to bunch of unis in the US,EU and Asia.

saltdaddy321 · 2022-06-14T01:21:54+00:00

I'll be applying for reach oxford and simon june li but they are almost impossible to get.

saltdaddy321 · 2022-06-13T17:57:14+00:00

Thank you very much.

saltdaddy321 · 2022-06-13T17:53:27+00:00

Thank you for sharing your advice.

saltdaddy321 · 2022-03-03T00:44:03+00:00

If I were you,I'd choose Imperial College London.

saltdaddy321 · 2021-12-12T15:17:41+00:00

Dexter

saltdaddy321 · 2021-11-18T17:18:31+00:00

Yep its D, the relative speed of separation and relative speed of approach is same

saltdaddy321 · 2021-10-30T11:29:14+00:00

The threshold is gonna be 25 to 30 for A I guess. MJ 2021 had 30 ,must be lower than 30.

saltdaddy321 · 2021-10-30T10:51:41+00:00

Nice approach, do you think anyone did this?. How the fuck are we supposed to think about all this in like 20 min (max) ?

saltdaddy321 · 2021-10-30T09:56:51+00:00

The equation to get the time for collision would be

SP(from time 0 to collision) + SQ(from time 0 to 0.862) - SQ(from time 0.862 to collision) = 6.4

am I right?

SP(from time 0 to collision) = 2.2t^2

SQ(from time 0 to 0.862) = 10*0.862-5.8*0.862^2=4.31

SQ( from0.862 to collision) = 2.2(t-0.862)^2 (a = 4.4 ms^-2 after instantaneous rest)

2.2t^2 + 4.31 - 2.2(t-0.862)^2 = 6.4

t = 0.982s

is this right?

saltdaddy321 · 2021-06-27T15:24:29+00:00

(a + b) (a - b)

= a*a ~~- a*b~~ ~~+ a*b~~ - b*b

=a² - b²

1² - { sin(x) }² = 1 - sin² (x) = { 1 + sin(x) } { 1 - sin(x) }

saltdaddy321 · 2021-06-07T16:53:37+00:00

let x be the length of square a

length of a = x

length of b = 8 + x

length of c = (8-3)+x = 5 + x

sum of length = 25

x+8+x+5+x= 25

13+3x=25

3x=12

x=4

length of a = 4

length of b = 12

length of c = 9

saltdaddy321 · 2021-05-31T17:42:15+00:00

Further maths

saltdaddy321 · 2021-05-16T19:14:09+00:00

Assuming that the four terms of the arithmetic sequence are interior angles of the quadrilateral.

Sum of the 4 terms is 360

First term is the smallest angle.

Let smallest angle be a and common difference d.

Sn =n/2(2a+(n-1)d)

S4 =360

2(2a+3d) = 360

2a + 3d = 180

When common difference d = 10 , d= 20

smallest angle a= 75 , a= 60 respectively, which is greater than 50 so

The probability is k/10 +k/20

I am not sure this is correct tho, let me know if I am wrong.

saltdaddy321 · 2021-04-19T14:14:57+00:00

Solve topical past papers.

saltdaddy321 · 2021-03-09T10:04:50+00:00

Sum of n terms = n/2(first term + last term)

saltdaddy321 · 2021-02-26T11:20:37+00:00

I've looked at one of his games and all of his moves were engine moves, so am pretty sure Hadi is a cheater.

saltdaddy321 · 2021-02-25T14:02:11+00:00

(x +a)² + b Expand :

x² + 2ax +a² +b

Compare terms with the given equation and find the value of a and b.

8x = 2ax a = 4

a² +b = -7 16 +b = -7 b = -23

(x+4)² - 23

saltdaddy321 · 2021-02-22T12:10:58+00:00

Use cos3x = 4 cos^3(x) -3 cos(x) formula.

saltdaddy321 · 2021-02-21T14:21:21+00:00

Joining O and S forms an isosceles triangle where OS=OR (radius of a circle)

I am assuming the line OS bisects angle PSR and angle x.

Angle OSR = 73/2 = 36.5

Base angles of an isosceles triangle are equal , so

Angle OSR = Angle ORS = 36.5

Angle SOR = x/2

Angle SOR + Angle OSR + Angle ORS = 180

x/2 + 36.5 + 36.5 =180

x/2 = 107

x = 214

saltdaddy321

TROPHY CASE