D. Budgeting time, avoiding stress, and prioritizing.
The solution would be like
this for this specific problem:
V^2 = 2AS = 2FS/M
V = sqrt(2FS/M) =
sqrt(2*105*.75/.087) = 44.52817783 = 42.5 mps
So the speed of the arrow as it leaves the bow
is 42.5 mps.
I am hoping that this answer has
satisfied your query and it will be able to help you in your endeavor, and if
you would like, feel free to ask another question.
Answer:
The mechanical advantage of the system is equal to 19.62
Explanation:
The ratio of the force produced by a machine to the force applied to it is called mechanical advantage. In other words it is the ratio of output force to the input force.
In this problem mass=200kg
applied force=100N
input force=100N
output force=
mechanical advantage 
It gives an idea about the efficiency of a mechanical device. It is indeed a measure of force amplification. In block and tackle system an assembly of ropes and pulleys is used to lift loads. When the moving block is supported by a greater number of rope sections the force amplification will be more.
Answer:
80,886 cm^3
Explanation:
v= lwh (also known as length × width × height)
v = 122cm × 51cm × 13cm
v = 80,886 cm^3
Answer:
i. The pressure of due to the water, <em>P</em>, is given according to the following equation;
P = ρ·g·h
Where;
ρ = The density of the water (a constant) = 997 kg/m³
g = The acceleration due to gravity = 9.81 m/s²
h = The height of the water (minimum h = h₁, maximum h = h₂)
The pressure is directly proportional to the water height, and we have;
The pressure, <em>P</em>, will be maximum when the water height, <em>h</em>, is maximum or h = h₂, which is the level DC
ii. The thrust = The force acting on the body = Pressure × Area
The maximum areas exposed to the water are on side AB and DC
However, the pressure at level DC, which is the location of the maximum pressure, is larger than the pressure at level AB, therefore, the maximum thrust will be at the level DC
Explanation:
