Qualification > Math
Maths/Physics help
astarmathsandphysics:
--- Quote from: omarsubei on February 11, 2009, 11:45:41 pm ---Please help me with this question and explain to me why the answer is what it is:
A force of magnitude F1 accelerates a body of mass m from rest to a speed v. A force of
magnitude F2 accelerates a body of mass 2m from rest to a speed 2v.
The ratio: (work done by F2)/(work done by F1) is:
A. 2.
B. 4.
C. 8.
D. 16.
Thanks
The accelerations are eqaul sinceF2/2m=F1/m therefore from v^2=u^"+2as, s=v^2/2a sobody 2 travels 4 times as far as body 1.
(work done by F2)/(work done by F1)=F2*4s/F1*s=4
--- End quote ---
astarmathsandphysics:
--- Quote from: jhonnyas on February 12, 2009, 07:03:18 am ---One question about Statistics
The height of certain plants are normally distributed. The plants are classified into three categories.
The shortest 12,92% are in cat a
The tallest 10,38% are in cat C
All other plants are in between r and t cm and are cat B
Given that the mean height is 6.84 cm and the standard deviation is 0.25 find t and r
PS> With explanations if possible, i m not an expert on statistics
--- End quote ---
I had this question a couple of days ago from one of my sudents.
For 12.92 % z=-1.13 so -1.13=(r-6.84)/025 so r=0.25*-1.13+6.84=6.5575
For10.38 we must do 1-.1038=0.8962 and z=1.26 s0 1.26=(t-6.84)/0.25 so t=1.26*0.25+6.84=7.155
omarsubei:
--- Quote from: astarmathsandphysics on February 12, 2009, 08:38:02 am ---
--- Quote from: omarsubei on February 11, 2009, 11:45:41 pm ---Please help me with this question and explain to me why the answer is what it is:
A force of magnitude F1 accelerates a body of mass m from rest to a speed v. A force of
magnitude F2 accelerates a body of mass 2m from rest to a speed 2v.
The ratio: (work done by F2)/(work done by F1) is:
A. 2.
B. 4.
C. 8.
D. 16.
Thanks
The accelerations are eqaul sinceF2/2m=F1/m therefore from v^2=u^"+2as, s=v^2/2a sobody 2 travels 4 times as far as body 1.
(work done by F2)/(work done by F1)=F2*4s/F1*s=4
--- End quote ---
--- End quote ---
That's the answer I came up with, but strangely the mark scheme says the answer is C (8). I was thinking maybe because it's asking about the "Work" done by F, and that there is distance involved (W=Fd), that the distance in F2 is double that of F1 because it's twice as fast?
astarmathsandphysics:
Can you copy and pasty the question
omarsubei:
I've attached it as a word document. Thank you.
Navigation
[0] Message Index
[#] Next page
[*] Previous page
Go to full version