Since the car is driven 125km west and then 65km north, we simply add the two values together to get 190km
Answer:
Part(a): The value of the spring constant is
.
Part(b): The work done by the variable force that stretches the collagen is
.
Explanation:
Part(a):
If '
' be the force constant and if due the application of a force '
' on the collagen '
' be it's increase in length, then from Hook's law
![F = k~\Delta l....................................................................(I)](https://tex.z-dn.net/?f=F%20%3D%20k~%5CDelta%20l....................................................................%28I%29)
Also, Young's modulus of a material is given by
![Y = \dfrac{F/A}{\Delta l/l}...............................................................(II)](https://tex.z-dn.net/?f=Y%20%3D%20%5Cdfrac%7BF%2FA%7D%7B%5CDelta%20l%2Fl%7D...............................................................%28II%29)
where '
' is the area of the material and '
' is the length.
Comparing equation (
) and (
) we can write
![&& Y = \dfrac{l~k}{A}\\&or,& k = \dfrac{Y~A}{l}\\&or,& k = \dfrac{Y~(\pi~r^{2})}{l}](https://tex.z-dn.net/?f=%26%26%20Y%20%3D%20%5Cdfrac%7Bl~k%7D%7BA%7D%5C%5C%26or%2C%26%20k%20%3D%20%5Cdfrac%7BY~A%7D%7Bl%7D%5C%5C%26or%2C%26%20k%20%3D%20%5Cdfrac%7BY~%28%5Cpi~r%5E%7B2%7D%29%7D%7Bl%7D)
Here, we have to consider only the circular surface of the collagen as force is applied only perpendicular to this surface.
Substituting the given values in equation (
), we have
![k = \dfrac{3.10 \times 10^{6}~N~m^{-2} \times \pi \times (0.00093)^{2}~m^{2}}{.027~m} = 3.11 \times 10^{2}~Kg~s^{-2}](https://tex.z-dn.net/?f=k%20%3D%20%5Cdfrac%7B3.10%20%5Ctimes%2010%5E%7B6%7D~N~m%5E%7B-2%7D%20%5Ctimes%20%5Cpi%20%5Ctimes%20%280.00093%29%5E%7B2%7D~m%5E%7B2%7D%7D%7B.027~m%7D%20%3D%203.11%20%5Ctimes%2010%5E%7B2%7D~Kg~s%5E%7B-2%7D)
Part(b):
We know the amount of work done (
) on the collagen is stored as a potential energy (
) within it. Now, the amount of work done by the variable force that stretches the collagen can be written as
![W = \dfrac{1}{2}k~x^{2} = \dfrac{(\dfrac{F}{k})^{2}k}{2} = \dfrac{F^{2}}{2~k}...................................(IV)](https://tex.z-dn.net/?f=W%20%3D%20%5Cdfrac%7B1%7D%7B2%7Dk~x%5E%7B2%7D%20%3D%20%5Cdfrac%7B%28%5Cdfrac%7BF%7D%7Bk%7D%29%5E%7B2%7Dk%7D%7B2%7D%20%3D%20%5Cdfrac%7BF%5E%7B2%7D%7D%7B2~k%7D...................................%28IV%29)
Substituting all the values, we can write
![W = \dfrac{(3.06 \times 10^{-2})^{2}~N^{2}}{2 \times 3.11 \times 10^{2}~Kg~s^{-2}} = 1.5 \times 10^{-6}~J](https://tex.z-dn.net/?f=W%20%3D%20%5Cdfrac%7B%283.06%20%5Ctimes%2010%5E%7B-2%7D%29%5E%7B2%7D~N%5E%7B2%7D%7D%7B2%20%5Ctimes%203.11%20%5Ctimes%2010%5E%7B2%7D~Kg~s%5E%7B-2%7D%7D%20%3D%201.5%20%5Ctimes%2010%5E%7B-6%7D~J)
Answer:
Cheetah cubs are in danger from predators like lions and hyenas which can track their prey by scent and so the mother and her cubs leave an area when their scent is too strong so that they are not hunted and the cubs survive.
Mother Cheetahs also train their cubs to hunt so that they may get food for themselves which will ensure their survival as well thus showing that both of these practices can impact on reproductive success.
The answer is : both lines should be the same lenght. Before towing 2 skiers, make sure the tow lines are of the same length if towing multiple skiers,and that they should wear a U.S. Coast Guard—approved life jacket (PFD) designed for water-skiing.