Answer:
Explanation:
The velocity at the inlet and exit of the control volume are same 
Calculate the inlet and exit velocity of water jet
The conservation of mass equation of steady flow
since inlet and exit velocity of water jet are equal so the inlet and exit cross section area of the jet is equal
The expression for thickness of the jet
R is the radius
t is the thickness of the jet
D_j is the diameter of the inlet jet
(b)
![R-x=\rho(AV_r)[-(V_i)+(V_c)\cos 60^o]\\\\=\rho(V_j+V_c)A[-(V_i+V_c)+(V_i+V_c)\cos 60^o]\\\\=\rho(V_j+V_c)(\frac{\pi}{4}D_j^2 )[V_i+V_c](\cos60^o-1)]](https://tex.z-dn.net/?f=R-x%3D%5Crho%28AV_r%29%5B-%28V_i%29%2B%28V_c%29%5Ccos%2060%5Eo%5D%5C%5C%5C%5C%3D%5Crho%28V_j%2BV_c%29A%5B-%28V_i%2BV_c%29%2B%28V_i%2BV_c%29%5Ccos%2060%5Eo%5D%5C%5C%5C%5C%3D%5Crho%28V_j%2BV_c%29%28%5Cfrac%7B%5Cpi%7D%7B4%7DD_j%5E2%20%29%5BV_i%2BV_c%5D%28%5Ccos60%5Eo-1%29%5D)
![R_x=[1000\times(44)\frac{\pi}{4} (10\times10^{-3})^2[(44)(\cos60^o-1)]]\\\\=-7603N](https://tex.z-dn.net/?f=R_x%3D%5B1000%5Ctimes%2844%29%5Cfrac%7B%5Cpi%7D%7B4%7D%20%2810%5Ctimes10%5E%7B-3%7D%29%5E2%5B%2844%29%28%5Ccos60%5Eo-1%29%5D%5D%5C%5C%5C%5C%3D-7603N)
The negative sign indicate that the direction of the force will be in opposite direction of our assumption
Therefore, the horizontal force is -7603N
Answer:
a. 1,188 N
b. 1.836
Explanation:
The computation is shown below:
a. For horizontal force, first we need to find out the circular path radius which is shown below:
As we know that
= 8m
Now the horizontal force is
where,
m = 66 kg
v = 12 m/s
and r = 8 m
Now placing these values to the above formula
So, the horizontal force is
= 1,188 N
b. Now the ratio of force to the weight of skater is
= 1.836
I’m deciding between E or D.. because for E it’s saying it’s changing ocean currents, but I’m thinking does that mean the earths precipitation or their climate. Best go with E, hope this helps
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.
The answer is D) <span>The molecular movement in the first rod transferred energy to the molecules in the second.</span>
