Answer:
Tissues that are damaged or injured.
Explanation:
Dystrophic calcification involves the deposition of calcium in soft tissues despite no disturbance in the calcium metabolism, and this is often seen at damaged tissues.
Examples of areas in the body where dystrophic calcification can occur include atherosclerotic plaques and damaged heart valves.
To solve this problem it is necessary to apply the concepts related to Newton's second law and its derived expressions for angular and linear movements.
Our values are given by,
![M_{cart} = 0.31kg\\m_{pulley} = 0.08kg\\r_{pulley} = 0.012m\\F = 1.1N\\](https://tex.z-dn.net/?f=M_%7Bcart%7D%20%3D%200.31kg%5C%5Cm_%7Bpulley%7D%20%3D%200.08kg%5C%5Cr_%7Bpulley%7D%20%3D%200.012m%5C%5CF%20%3D%201.1N%5C%5C)
If we carry out summation of Torques on the pulley we will have to,
![F_2*d-F_1*d = I \alpha](https://tex.z-dn.net/?f=F_2%2Ad-F_1%2Ad%20%3D%20I%20%5Calpha)
Where,
I = Inertia moment
Angular acceleration, which is equal in linear terms to a/r (acceleration and radius)
The moment of inertia for this object is given as
![I = \frac{1}{2} mr^2](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20mr%5E2)
Replacing this equations we have know that
![(F_2 - F_1)d = (\frac{1}{2}(m_{pulley})r^2) (\frac{a}{r})](https://tex.z-dn.net/?f=%28F_2%20-%20F_1%29d%20%3D%20%28%5Cfrac%7B1%7D%7B2%7D%28m_%7Bpulley%7D%29r%5E2%29%20%28%5Cfrac%7Ba%7D%7Br%7D%29)
![F_2 - F_1 = \frac{1}{2}m_{pulley} \frac{F_1}{M_{cart}}](https://tex.z-dn.net/?f=F_2%20-%20F_1%20%3D%20%5Cfrac%7B1%7D%7B2%7Dm_%7Bpulley%7D%20%5Cfrac%7BF_1%7D%7BM_%7Bcart%7D%7D)
![F_2 = (1+\frac{1}{2}(\frac{m_{pulley}}{M_{cart}}))F_1](https://tex.z-dn.net/?f=F_2%20%3D%20%281%2B%5Cfrac%7B1%7D%7B2%7D%28%5Cfrac%7Bm_%7Bpulley%7D%7D%7BM_%7Bcart%7D%7D%29%29F_1)
Or
![F_1 = \frac{F_2}{(1+\frac{1}{2}(\frac{m_{pulley}}{M_{cart}}))}](https://tex.z-dn.net/?f=F_1%20%3D%20%5Cfrac%7BF_2%7D%7B%281%2B%5Cfrac%7B1%7D%7B2%7D%28%5Cfrac%7Bm_%7Bpulley%7D%7D%7BM_%7Bcart%7D%7D%29%29%7D)
Replacing our values we have that
![F_1 = \frac{1.1}{(1 + (0.5)(\frac{0.08}{0.31}))}](https://tex.z-dn.net/?f=F_1%20%3D%20%5Cfrac%7B1.1%7D%7B%281%20%2B%20%280.5%29%28%5Cfrac%7B0.08%7D%7B0.31%7D%29%29%7D)
![F_1 = 0.974 N](https://tex.z-dn.net/?f=F_1%20%3D%200.974%20N)
Therefore the tension in the string between the pulley and the cart is 0.974 N
Answer:
An effect whereby a mass moving in a rotating system experienced a force acting perpendicular to the direction of motion and to the axis of rotation.
The shell's vertical position
at time
is given by
![y=25.0\,\mathrm m-\dfrac g2t^2](https://tex.z-dn.net/?f=y%3D25.0%5C%2C%5Cmathrm%20m-%5Cdfrac%20g2t%5E2)
where
is the acceleration due to gravity. The shell hits the water when
, so the time it takes for that to happen is
![0=25.0\,\mathrm m-\dfrac g2t^2\implies t=2.26\,\mathrm s](https://tex.z-dn.net/?f=0%3D25.0%5C%2C%5Cmathrm%20m-%5Cdfrac%20g2t%5E2%5Cimplies%20t%3D2.26%5C%2C%5Cmathrm%20s)
For the shell to reach a height of
, it must travel for a time
such that
![5.00\,\mathrm m=25.0\,\mathrm m-\dfrac g2t^2\implies t=2.02\,\mathrm s](https://tex.z-dn.net/?f=5.00%5C%2C%5Cmathrm%20m%3D25.0%5C%2C%5Cmathrm%20m-%5Cdfrac%20g2t%5E2%5Cimplies%20t%3D2.02%5C%2C%5Cmathrm%20s)
The shell's horizontal position
is given by
![x=\left(75.0\,\dfrac{\mathrm m}{\mathrm s}\right)t](https://tex.z-dn.net/?f=x%3D%5Cleft%2875.0%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%7D%5Cright%29t)
so that after 2.02 seconds, the warship should be at a distance
![x=\left(75.0\,\dfrac{\mathrm m}{\mathrm s}\right)(2.02\,\mathrm s)=153\,\mathrm m](https://tex.z-dn.net/?f=x%3D%5Cleft%2875.0%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%7D%5Cright%29%282.02%5C%2C%5Cmathrm%20s%29%3D153%5C%2C%5Cmathrm%20m)
away from the pirate ship in order to hit it at the desired height.
To solve this problem we must resort to the definition of Normal Force, in which, it is the component of a contact force that is perpendicular to the surface that an object contacts.
![F = mg](https://tex.z-dn.net/?f=F%20%3D%20mg)
where,
m=mass
g= Gravitational Acceleration
Normal force that the force exerts on the ladder is the weight of the ladder, then
F=mg
F=18*9.8
F=176.58N
Therefore normal force the floor exerts on the ladder is 176.58N