Answer:
<em>Velocity is the rate at which the position changes</em>
<em>Velocity is the rate at which the position changesWhy do we need</em>
<em>Velocity is the rate at which the position changesWhy do we needVectors make it convenient to handle quantities going in different directions</em><em>.</em><em>.</em><em> </em>
Explanation:
Thank you! 
 
        
                    
             
        
        
        
Answer:
h = 13.06 m
Explanation:
Given:
- Specific gravity of gasoline S.G = 0.739
- Density of water p_w = 997 kg/m^3
- The atmosphere pressure P_o = 101.325 KPa
- The change in height of the liquid is h m
Find:
How high would the level be in a gasoline barometer at normal atmospheric pressure?
Solution:
- When we consider a barometer setup. We dip the open mouth of an inverted test tube into a pool of fluid. Due to the pressure acting on the free surface of the pool, the fluid starts to rise into the test-tube to a height h.
- The relation with the pressure acting on the free surface and the height to which the fluid travels depends on the density of the fluid and gravitational acceleration as follows:
                                          P = S.G*p_w*g*h
Where,                              h = P / S.G*p_w*g
- Input the values given:
                                          h = 101.325 KPa / 0.739*9.81*997
                                          h = 13.06 m
- Hence, the gasoline will rise up to the height of 13.06 m under normal atmospheric conditions at sea level. 
 
        
             
        
        
        
The total momentum should come out to be  <span>2.0 x 10^4 kilogram meters/second </span>
        
                    
             
        
        
        
A force of 43.8 N is required to stretch the spring a distance of 15.5 cm = 0.155 m, so the spring constant <em>k</em> is
43.8 N = <em>k</em> (0.155 m)   ==>   <em>k</em> = (43.8 N) / (0.155 m) ≈ 283 N/m
The total work done on the spring to stretch it to 15.5 cm from equilibrium is
1/2 (283 N/m) (0.155 m)² ≈ 3.39 J
The total work needed to stretch the spring to 15.5 cm + 10.4 cm = 25.9 cm = 0.259 m from equilibrium would be
1/2 (283 N/m) (0.259 m)² ≈ 9.48 J
Then the additional work needed to stretch the spring 10.4 cm further is the difference, about 6.08 J.