Answer:
The total momentum of the system before the collision is 0.0325 kg-m/s due east direction.
Explanation:
Given that,
Mass of the cart, m = 250 g = 0.25 kg
Initial velocity of the cart, u = 0.31 m/s (due right)
Mass of another cart, m' = 500 g = 0.5 kg
Initial velocity of the another cart u' = -0.22 m/s (due left)
Let p is the total momentum of the system before the collision. It is given by :
So, the total momentum of the system before the collision is 0.0325 kg-m/s due east direction.
Evaporation (or another word to use is water vapor.)
Add then divide the hint clearly backs it up two so yeahh
Answer:
The work done to get you safely away from the test is 2.47 X 10⁴ J.
Explanation:
Given;
length of the rope, L = 70 ft
mass per unit length of the rope, μ = 2 lb/ft
your mass, W = 120 lbs
mass of the 70 ft rope = 2 lb/ft x 70 ft
= 140 lbs.
Total mass to be pulled to the helicopter, M = 120 lbs + 140 lbs
= 260 lbs
The work done is calculated from work-energy theorem as follows;
W = Mgh
where;
g is acceleration due gravity = 32.17 ft/s²
h is height the total mass is raised = length of the rope = 70 ft
W = 260 Lb x 32.17 ft/s² x 70 ft
W = 585494 lb.ft²/s²
1 lb.ft²/s² = 0.0421 J
W = 585494 lb.ft²/s² = 2.47 X 10⁴ J.
Therefore, the work done to get you safely away from the test is 2.47 X 10⁴ J.
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