When developing an experimental design, the action that would improve the quality of the results is to ensure that it answers a question about cause and effect.
<h3>What is experimental design?</h3>
Experimental design is a concept used to organize, conduct, and interpret results of experiments in an efficient way, making sure that as much useful information as possible is obtained by performing a small number of trials.
Thus, when developing an experimental design, the action that would improve the quality of the results is to ensure that it answers a question about cause and effect.
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Given parameters:
Initial velocity of Coin = 0m/s
Time taken before coin hits ground = 5.7s
Unknown:
Final velocity of the coin = ?
Velocity is displacement with time. To solve this problem, we have to apply one of the equations of motion.
The fitting one of them here is shown below;
V = U + gt
where;
V is the final velocity
U is the initial velocity
g is the acceleration due to gravity
t is the time taken
Here we use positive value of acceleration due to gravity because the coin is falling with the effect of acceleration and not against it.
Now input the parameters and solve;
V = 0 + 9.81 x 5.7
V = 55.917m/s
Therefore, the final velocity is 55.917m/s.
Answer:
Yes it is
Explanation:
the sum moles at the left side equals the sum of moles at the right side
Answer:
W = ½ m v²
Explanation:
In this exercise we must solve it in parts, in a first part we use the conservation of the moment to find the speed after the separation
We define the system formed by the two parts of the rocket, therefore the forces during internal separation and the moment are conserved
initial instant. before separation
p₀ = m v
final attempt. after separation
= m /2 0 + m /2 v_{f}
p₀ = p_{f}
m v = m /2 
v_{f}= 2 v
this is the speed of the second part of the ship
now we can use the relation of work and energy, which establishes that the work is initial to the variation of the kinetic energy of the body
initial energy
K₀ = ½ m v²
final energy
= ½ m/2 0 + ½ m/2 v_{f}²
K_{f} = ¼ m (2v)²
K_{f} = m v²
the expression for work is
W = ΔK = K_{f} - K₀
W = m v² - ½ m v²
W = ½ m v²
The answer should be speed hope this helps