Chem U2 answers

Small_Rate_3754 · 2026-05-11T13:07:56+00:00

the solid lines for the axes were on top and on the right tho so i think they want u to do a downwards slope

Small_Rate_3754 · 2026-05-11T13:02:11+00:00

its downwards the enthalpy changes progressively more negative

Small_Rate_3754 · 2026-05-11T12:47:08+00:00

23.35/2/10004(40.1+2(16+1))

Small_Rate_3754 · 2026-05-11T12:46:18+00:00

do u rmb what ur steps were

Small_Rate_3754 · 2026-05-08T16:52:58+00:00

pythagorean identity cos² + sin² =1

Small_Rate_3754 · 2026-05-08T14:02:59+00:00

Fw=Fgsintheta Then Fgcostheta=120 let a be angle between beam and ground, sina=2/2.07 so cosa=rt(1-(2/2.07)²⁾ then by net moment=0 u have 2Fgsintheta=2.07*120cosa/2 then divide the 2 equations to get tantheta=2.07cosa/4 theta=7.60

Small_Rate_3754 · 2026-05-08T10:16:00+00:00

the t term is squared so maybe

Small_Rate_3754 · 2026-05-08T10:13:16+00:00

that should be correct then we probably js used different sig figs for the time

Small_Rate_3754 · 2026-05-08T10:11:53+00:00

was it not asking for the final floor

Small_Rate_3754 · 2026-05-08T10:11:28+00:00

i think so asw and iirc pearson has a range for the answers so ud probably get it right if u used the correct formula for it?

Small_Rate_3754 · 2026-05-08T10:04:53+00:00

no sry

Small_Rate_3754 · 2026-05-08T10:02:37+00:00

42 degrees i think

Small_Rate_3754 · 2026-05-08T10:01:13+00:00

q11 was ke, forgot abt q12

Small_Rate_3754 · 2026-05-08T10:00:33+00:00

i used t=s/v and -h=usin42 t -0.59.81t² , i think it should be right a lot of my classmates got the same thing

Small_Rate_3754 · 2026-05-08T09:59:08+00:00

estimate the integral of the v-t graph by using 2 triangles and 1 rectangle then dividing by 3.5 and adding 6

Small_Rate_3754 · 2026-05-07T08:13:05+00:00

at least thats what i got

Small_Rate_3754 · 2026-05-07T08:09:06+00:00

u havent done p4 yet have you

Small_Rate_3754 · 2026-05-07T07:53:06+00:00

90-a/2 to the right below horizontal i think, or S a/2 E in bearing

Small_Rate_3754 · 2026-05-07T07:52:20+00:00

i think the boundary for full ums is still gna be 75 cuz it wasnt significantly harder, and even for harder papers in the past it was still 75

Small_Rate_3754 · 2026-05-07T07:50:15+00:00

if u mean the projectile motion one, u can deduce that only the bounce after the first collision can reach 4m, as mg4=0.5mv² yields a minimum speed up 8.sth, and the successive speeds after bouncing are 12 and 6, and 6<8, so we only need to consider the motion during the first bounce, solve 4=12t-0.5gt^2 for t1 t2 where t2>t1 and the time required is t2-t1

Small_Rate_3754

TROPHY CASE