Author Topic: Maths/Physics help  (Read 23686 times)

Offline lolo

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Re: Maths/Physics help
« Reply #15 on: February 05, 2009, 12:11:37 pm »
i have another question plz:

there were 1240 tickets sold for a concert. Of these, 780 were sold at $24.50 each, while the remainder were sold at a reduction of 20%. Find the total money taken in ticket sales.

thanks :)

Offline astarmathsandphysics

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Re: Maths/Physics help
« Reply #16 on: February 05, 2009, 02:33:12 pm »
i have another question plz:

there were 1240 tickets sold for a concert. Of these, 780 were sold at $24.50 each, while the remainder were sold at a reduction of 20%. Find the total money taken in ticket sales.

thanks :)
(780x24.5)+ 0.8x(1240-780)x24.5=28136

Offline lolo

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Re: Maths/Physics help
« Reply #17 on: February 08, 2009, 05:50:09 am »
thanks .. bt i reli dnt get what u did  ???.. cn u plz explain
thanks :)

Offline astarmathsandphysics

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Re: Maths/Physics help
« Reply #18 on: February 08, 2009, 07:35:13 am »
780 sold at 24.50 so revenue =780x24.5=19110

The remainder have a discount of 20% iethey cost 24.50x0.8=19.60. There are 1240-780=460 of these so revenue =19.60x460=9016

Total revenue =19110+9016=28126

Offline omarsubei

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Re: Maths/Physics help
« Reply #19 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

Offline astarmathsandphysics

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Re: Maths/Physics help
« Reply #20 on: February 12, 2009, 08:38:02 am »
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


Offline astarmathsandphysics

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Re: Maths/Physics help
« Reply #21 on: February 12, 2009, 08:45:15 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
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

Offline omarsubei

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Re: Maths/Physics help
« Reply #22 on: February 12, 2009, 10:14:17 am »
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


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?

Offline astarmathsandphysics

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Re: Maths/Physics help
« Reply #23 on: February 12, 2009, 10:49:39 am »
Can you copy and pasty the question

Offline omarsubei

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Re: Maths/Physics help
« Reply #24 on: February 12, 2009, 11:09:47 am »
I've attached it as a word document. Thank you.

Offline astarmathsandphysics

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Re: Maths/Physics help
« Reply #25 on: February 12, 2009, 11:19:22 am »
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


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?
0.5(2M)V^2/0.mvv^"=8 THEquestion is badlty worded

Offline omarsubei

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Re: Maths/Physics help
« Reply #26 on: February 12, 2009, 11:31:28 am »
Sorry I don't understand. Could you please explain.

Offline astarmathsandphysics

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Re: Maths/Physics help
« Reply #27 on: February 12, 2009, 11:32:49 am »
f2 is not twicw F1

Offline astarmathsandphysics

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Re: Maths/Physics help
« Reply #28 on: February 12, 2009, 11:34:03 am »
u hve to work out the ratiof2/f1 first.

Offline omarsubei

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Re: Maths/Physics help
« Reply #29 on: February 12, 2009, 11:38:24 am »
You're making me feel even more confused. :-\
The ratio of F2/F1 is 4. How does the "work"  make that 8?