Answer:
it means it has pressure of 13.6 CM mercury on an area
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.
momentum= mass × velocity = 0.141kg×1.33m/s= 0.18753kg m/s = 0.188kg m/s (3s.f.)
r(t) models the water flow rate, so the total amount of water that has flowed out of the tank can be calculated by integrating r(t) with respect to time t on the interval t = [0, 35]min
∫r(t)dt, t = [0, 35]
= ∫(300-6t)dt, t = [0, 35]
= 300t-3t², t = [0, 35]
= 300(35) - 3(35)² - 300(0) + 3(0)²
= 6825 liters