To be able to classify which species is which
Answer:
u=36.8m/s
Explanation:
because of the acceleration is a constant acceleration we can use one of the "SUVAT" equations
u^2=v^2-2ā*s. where:
u^2 stands for intial velocity
v^2 stands for final velocity
since the cougar skidded to a complete stop the final velocity is zero.
u^2=v^2-2ā*s
u^2=(0)^2 -2(-2.87 m/s^2)*236 m
u^2=0+5.74m/s^2* 236m
u^2=1354.64m^2/s^2
u=√1354.64m^2/s^2
u=36.8m/s (approximate value)
when ever the acceleration is constant you can use one of the following equation to find the required value.
1. v = u + at. (no s)
2. s= 1/2(u+v)t. (no ā)
3. s=ut + 1/2at^2. ( no v)
4. v^2=u^2 + 2āS. (no t). 5. s= vt - 1/2at^2. (no u)
Answer:
The bottom of the sea is 25 m below sea level.
Explanation:
Given data
Mass = 6.1 × 
We know that Buoyant force on the tank is equal to gravity force of the tank.
1020 × 
= 6.1 × 
= 598039.21 
We know that
= W × L × H
598039.21 = 300 × 80 × H
H = 25 m
Therefore the bottom of the sea is 25 m below sea level.
The question is asking to calculate the object's speed v1, v2, v3 at the certain time is the given of the problem, in my calculation, I would say that the speed would be 2m/s, 1.5m/s, 0.22m/s. I hope you are satisfied with my answer and feel free to ask for more if you have question and further clarification
