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[synthesisFR]

 

[Luukas.2]

Taking to the right and upwards as the directions for i and j:

The weight force is W = -50g j

The tensions are F = F[(cos 53) i + sin 53) j] and T = T[(-cos 40) i + (sin 40) j], where F and T are the magnitudes of the two tensions, respectively.

Since the object is stationary, F + T + W = 0

Hence: Fcos 53 - Tcos 40 = 0 and Fsin 53 + Tsin 40 - 50g = 0.

 

[synthesisFR]

Taking to the right and upwards as the directions for i and j:

The weight force is W = -50g j

The tensions are F = F[(cos 53) i + sin 53) j] and T = T[(-cos 40) i + (sin 40) j], where F and T are the magnitudes of the two tensions, respectively.

Since the object is stationary, F + T + W = 0

Hence: Fcos 53 - Tcos 40 = 0 and Fsin 53 + Tsin 40 - 50g = 0.

i couldn't solve that last line..


why is x<-2 discarded bc can't it move back. idk

 

[synthesisFR]

idk how to do

 

[synthesisFR]

how to do ii

 

[Luukas.2]

i couldn't solve that last line..

View attachment 40462
why is x<-2 discarded bc can't it move back. idk

Look at the velocity equation... it is undefined for x = 0 and x = 2, so the particle is always trapped in x > 2, 0 < x < 2, or x < 0.

Given x = 1 initially, the domain of movement is no bigger than 0 < x < 2

 

[synthesisFR]

i dont get where the 45, 43.8 come from

 

[Trippzy]

View attachment 40460View attachment 40461

for (c) i) is the answer negatively skewed

 

[synthesisFR]

yeah but how

 

[Luukas.2]

i couldn't solve that last line..

View attachment 40462
why is x<-2 discarded bc can't it move back. idk

Note that a much simpler approach is to sketch



as both and are easily sketched and you can then add them to get y = f(x). The question tells you that v = 0 at x = 1 and so the sketch of v² v. x is simply the part of y = f(x) where .

This sketch makes obvious that the motion is restricted to without solving a single inequality.

 

[WeiWeiMan]

yeah but how

p=0.23 which is quite a big smaller than 0.5 so the things are bunched to the left

 

[Trippzy]

yeah but how

the probability is 0.23 for the students who do ext 1
for a binomial distribution to be negatively skewed, the probability must be less than 0.5. so in this case 0.23 < 0.5 = negatively skewed

conversely for positive skewed it must be X > 0.5

don't ask me how it workers idk i just know that's what u do

 

[synthesisFR]

Where we meant toearn that
I fee like I missed out on content bruh

 

[carrotsss]

Where we meant toearn that
I fee like I missed out on content bruh

the positive/negative skews? yeah

 

[synthesisFR]

am i dum or is this wrong

 

[carrotsss]

View attachment 40475
am i dum or is this wrong

-(3pi + 3pi/4) = -(2pi + 7pi/4) = -7pi/4 = pi/4 (reducing to principal argument)

 

[synthesisFR]

the sqrt 2

 

[mmmmmmmmaaaaaaa]

the sqrt 2

i think its wrong