<span>The first stage in the Gas model of stress is alarm and
mobilization. So the correct option in regards to the given question is option “d”.
Hans Selye is the person that evolved this model and he has explained this
model in complete details. He has broken
down his model into three stages. The first stage involves alarm and
mobilization. The second stage includes resistance. The third and the final
stage include the exhaustion stage. These are the stages that an organism goes
through to restore back the balance when stress is exerted from outside. </span>
Answer:
The velocity of the ball before it hits the ground is 381.2 m/s
Explanation:
Given;
time taken to reach the ground, t = 38.9 s
The height of fall is given by;
h = ¹/₂gt²
h = ¹/₂(9.8)(38.9)²
h = 7414.73 m
The velocity of the ball before it hits the ground is given as;
v² = u² + 2gh
where;
u is the initial velocity of the on the root = 0
v is the final velocity of the ball before it hits the ground
v² = 2gh
v = √2gh
v = √(2 x 9.8 x 7414.73 )
v = 381.2 m/s
Therefore, the velocity of the ball before it hits the ground is 381.2 m/s
<span>R = rate of flow = 0.370 L/s
H = height = 2.9 m
T= time = 3.9 s
V = velocity of water when it hits the bucket = sqrt(2gh) = sqrt(2 x 9.8 x 2.9) =7.539 m/s2
G value = 9.8 m/s2
Wb = weight of bucket = 0.690 kg x 9.8 m/s2 = 6.762 N
Wa = weight of accumulated water after 3.9 s
Fi = force of impact of water on the bucket
S = reading on the scale = Wa + Wb + Fi
mass of water accumulated after 3.9 s = R x T = 0.370 x 3.9 = 1.443 L = 1.443 kg
Therefore, Wa = 1.443 x 9.8 = 14.1414 N
Fi = rate of change of momentum at the impact point = R x V (because R = dm/dt)
= 0.37 x 7.539 = 2.78943 N
S = 14.1414 + 6.762 + 2.78943 = 23.692 N</span>
7.17m/s glad I could help
