D = v^2 / 2ug
d= 3.5^2 / 0,15 x 9.8 m/s^2
the answer should be around 4.2m
hope this helps
Answer:
Using lighter material in car construction, improving energy efficiency by enhancing engine design or replacing the engine with more efficient technologies.
Explanation:
Using lighter materials in the car construction, reducing the potential energy required to accelerate and to move the car, as well as energy losses due to rolling friction. There is evidence of such benefits by replacing steel and aluminium parts with components made of composite materials.
Improving the design of internal combustion engines to minimize energy losses and accordingly, improving energy efficiency. A more radical approach is replacing internal combustion engines with electric engines, which offer higher efficiencies. Such conclusions can be easily inferred from model based on Work-Energy Theorem and Principle of Energy Conservation:

Answer:
2.43J
Explanation:
Given parameters:
Mass of the arrow = 0.155kg
Velocity = 31.4m /s
Unknown:
Kinetic energy when it leaves the bow = ?
Solution:
The kinetic energy of a body is the energy in motion of the body;
it can be derived using the expression below:
K.E =
m v²
m is the mass
v is the velocity
Solve for K.E;
K.E =
x 0.155 x 31.4 = 2.43J
Answer:
a) 
b) 
Explanation:
Given:
height of water in one arm of the u-tube, 
a)
Gauge pressure at the water-mercury interface,:

we've the density of the water 


b)
Now the same pressure is balanced by the mercury column in the other arm of the tube:



<u>Now the difference in the column is :</u>


