Answer:
Acceleration of the object is
.
Explanation:
It is given that, the position of the object is given by :
![r=[2\ m+(5\ m/s)t]i+[3\ m-(2\ m/s^2)t^2]j](https://tex.z-dn.net/?f=r%3D%5B2%5C%20m%2B%285%5C%20m%2Fs%29t%5Di%2B%5B3%5C%20m-%282%5C%20m%2Fs%5E2%29t%5E2%5Dj)
Velocity of the object, 
Acceleration of the object is given by :

![a=\dfrac{d^2}{dt^2}([2\ m+(5\ m/s)t]i+[3\ m-(2\ m/s^2)t^2]j)](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7Bd%5E2%7D%7Bdt%5E2%7D%28%5B2%5C%20m%2B%285%5C%20m%2Fs%29t%5Di%2B%5B3%5C%20m-%282%5C%20m%2Fs%5E2%29t%5E2%5Dj%29)
Using the property of differentiation, we get :

So, the magnitude of the acceleration of the object at time t = 2.00 s is
. Hence, this is the required solution.
It's Z.
Without any force acting on it an object travels in a straight line.
In order to bend away from a straight line the object needs a force acting on it.
In order to move along a circle, the force on the object points toward the center of the circle. It's called the centripetal force.
Since the object's direction is changing it has acceleration.
The acceleration points toward the center of the circle.
Answer:
x2 = 0.99
Explanation:
from superheated water table
at pressure p1 = 0.6MPa and temperature 200 degree celcius
h1 = 2850.6 kJ/kg
From energy equation we have following relation



![2850.6 + [\frac{50^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}] = h2 +[ \frac{600^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}]](https://tex.z-dn.net/?f=2850.6%20%2B%20%5B%5Cfrac%7B50%5E2%7D%7B2%7D%20%2A%20%5Cfrac%7B1%20kJ%2Fkg%7D%7B1000%20m%5E2%2FS%5E2%7D%5D%20%3D%20h2%20%2B%5B%20%5Cfrac%7B600%5E2%7D%7B2%7D%20%2A%20%5Cfrac%7B1%20kJ%2Fkg%7D%7B1000%20m%5E2%2FS%5E2%7D%5D)
h2 = 2671.85 kJ/kg
from superheated water table
at pressure p2 = 0.15MPa
specific enthalpy of fluid hf = 467.13 kJ/kg
enthalpy change hfg = 2226.0 kJ/kg
specific enthalpy of the saturated gas hg = 2693.1 kJ/kg
as it can be seen from above value hf>h2>hg, so phase 2 is two phase region. so we have
quality of steam x2
h2 = hf + x2(hfg)
2671.85 = 467.13 +x2*2226.0
x2 = 0.99
The refraction of light at the surface of water makes ponds look shallower then they really are