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Physics question! Plz answerr! VERY IMPORTANT
One species of dinosaur had a mass of about 1500 kg, as it walked, could support its whole weight on a single leg bone of length 75cm. If the breaking stress of the bone is 1.8 x 10(the 10 is to the power of 7) Pa then the minimum diameter the leg could have been to support the animal without fracture is?
I need the answer and ALL the workings out....
It is very important and is needed ASAP
Thankss Btw its either a,b,c or d
(a)1.6cm
(b)2.8cm
(c)3.2cm
(d)4.8cm
I need the answer and ALL the workings out....
It is very important and is needed ASAP
Thankss Btw its either a,b,c or d
(a)1.6cm
(b)2.8cm
(c)3.2cm
(d)4.8cm
Answers
Best Answer
No best answer has yet been selected by gotchaeme. Once a best answer has been selected, it will be shown here.
For more on marking an answer as the "Best Answer", please visit our FAQ.if pressure = force/area
then area = force/pressure = weight/pressure
= (1500 x 10)/18000000
= 8.33 x 10^(-4) metres squared
using 10 metres/second squared as acceleration due to gravity
area = pi x r x r
r = sqrt(area/pi) = sqrt(2.65) = 0.016 m, i.e 1.6 cm
The length of the leg bone doesn't actually matter here, as the weight is assumed to be acting straight downwards (and the bone is assumed cylindrical like a pipe), so no matter whether it is short or long the breaking stress is the same (stress = force/area in Youngs Modulus (stress/strain where strain = extension/original length)).
It matter to the dinosaur though, as it could be a long way to fall or a long distance along which blood has to be pumped!
Hope that makes some sense
then area = force/pressure = weight/pressure
= (1500 x 10)/18000000
= 8.33 x 10^(-4) metres squared
using 10 metres/second squared as acceleration due to gravity
area = pi x r x r
r = sqrt(area/pi) = sqrt(2.65) = 0.016 m, i.e 1.6 cm
The length of the leg bone doesn't actually matter here, as the weight is assumed to be acting straight downwards (and the bone is assumed cylindrical like a pipe), so no matter whether it is short or long the breaking stress is the same (stress = force/area in Youngs Modulus (stress/strain where strain = extension/original length)).
It matter to the dinosaur though, as it could be a long way to fall or a long distance along which blood has to be pumped!
Hope that makes some sense
that's exactly what I've done too!
pressure = force/area = weight/(pi x r squared)
r squared = weight/(pi x pressure)
r = sqrt(weight/(pi x pressure))
It may be easier for you step by step. First calculate the area as a numerical number as you have been given the inputs to do so. Then you know that this area = pi x r squared.
so area/pi = r squared
r = squareroot of (area/pi)
pressure = force/area = weight/(pi x r squared)
r squared = weight/(pi x pressure)
r = sqrt(weight/(pi x pressure))
It may be easier for you step by step. First calculate the area as a numerical number as you have been given the inputs to do so. Then you know that this area = pi x r squared.
so area/pi = r squared
r = squareroot of (area/pi)
always try and break the problem down into its simplest form and constituent parts. Write down what you know from the question, this often helps you to remember a formula in an exam, then try to logically reach your answer.
Examining units is also a good idea in physics; if you know the units of the answer, you can often work out what has to be multiplied by what (or whatever) to reach it.
A diagram is always good too!
Examining units is also a good idea in physics; if you know the units of the answer, you can often work out what has to be multiplied by what (or whatever) to reach it.
A diagram is always good too!
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