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AS Core 1 Question Help!

Hi. Please could I have some help on 10,11 and 12? (In image):

image.jpg

Thanks in advance! :-)
Reply 1
Avatar for Yusufff
Yusufff
OP
10
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Posted from TSR Mobile
10. l = 2.5

Working out: 2l = 1.5 + (l+1)
l = 1.5 + 1
l = 2.5

I don't quite understand the wording for 11, sorry.

12.
A. Answer is in attachment (plus working) (sorry, other answer is 3/5, not 5/3...)
B. The ball will be at 4.5m above the ground at two times, when it is going up and again when coming down.

WP_20150924_004.jpg

EDIT:
Got numbers upside down...
(edited 8 years ago)
Reply 3
Avatar for RMNDK
RMNDK
3
Original post by Yusufff
Hi. Please could I have some help on 10,11 and 12? (In image):

image.jpg

Thanks in advance! :-)


Q11

The surface of the disc is the surface area of a cylinder.
SA of a Cylinder = 2πr2 + πdl
However, thickness is negligible so therefore, SA = 2πr2

The surface area of a ball bearing is the surface area of a sphere and is even given to you...
SA of a Sphere = 4πr2

You know what r is for the sphere.. it's just r.
But the radius of the disc is r-1.
So the SA of the disc is SA = 2π(r-1)2

Total SA = 132π
Total SA = SA of disc + SA of sphere

THEREFORE: 132π = 2π(r-1)2 + 4πr2
132π = 2πr2 + 2π - 4πr + 4πr2

I've helped you way too much. Solve that quadratic, if you can't, learn your basics
Reply 4
Avatar for Xsk
Xsk
4
10) @Scienceisgood ??
All you need to do is work out the area by creating a quadratic.
2l(2l) + 1.5(l + 1) = 22.5
4l^2 + 1.5l +1.5 - 22.5 = 0
4l^2 +1.5l - 21 = 0
Use quadratic formula;
l = 2.11 or l = -2.49
-2.49 isn't possible because l is a length, therefore l = 2.11m

(I have the textbook too, answer at the back says 2.11m :P)
Reply 5
Avatar for Yusufff
Yusufff
OP
10
Thank you very much for your responses. All helped me very well in seeing where I went wrong.

Original post by Scienceisgood
10. l = 2.5

Working out: 2l = 1.5 + (l+1)
l = 1.5 + 1
l = 2.5

I don't quite understand the wording for 11, sorry.

12.
A. Answer is in attachment (plus working) (sorry, other answer is 3/5, not 5/3...)
B. The ball will be at 4.5m above the ground at two times, when it is going up and again when coming down.

WP_20150924_004.jpg

EDIT:
Got numbers upside down...


Original post by RMNDK
Q11

The surface of the disc is the surface area of a cylinder.
SA of a Cylinder = 2πr2 + πdl
However, thickness is negligible so therefore, SA = 2πr2

The surface area of a ball bearing is the surface area of a sphere and is even given to you...
SA of a Sphere = 4πr2

You know what r is for the sphere.. it's just r.
But the radius of the disc is r-1.
So the SA of the disc is SA = 2π(r-1)2

Total SA = 132π
Total SA = SA of disc + SA of sphere

THEREFORE: 132π = 2π(r-1)2 + 4πr2
132π = 2πr2 + 2π - 4πr + 4πr2

I've helped you way too much. Solve that quadratic, if you can't, learn your basics


Original post by Xsk
10) @Scienceisgood ??
All you need to do is work out the area by creating a quadratic.
2l(2l) + 1.5(l + 1) = 22.5
4l^2 + 1.5l +1.5 - 22.5 = 0
4l^2 +1.5l - 21 = 0
Use quadratic formula;
l = 2.11 or l = -2.49
-2.49 isn't possible because l is a length, therefore l = 2.11m

(I have the textbook too, answer at the back says 2.11m :P)

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