Its B. I know because I just took the test in K12 so your welcome EJ.
Answer:
Δe=0.578 kJ/kg
Explanation:
Given data
Velocity v₁=0 m/s
Velocity v₂=34 m/s
to find
Specific energy change Δe
Solution
The specific energy change is simply determined from change in velocity
Δe=(v₂²-v₁²)/2
Put the given values to find the specific energy change

Δe=0.578 kJ/kg
Answer:
1.8m
Explanation:
Let the Elastics of the steel ASTM-36 
The strain of the bar when subjected to 150 MPa is

Therefore, if the bar elongates by 1.35 mm, then the original length L would be:

or 1.8m
Power is defined as rate of work done which means it is work done in 1 second of time
Now it is given that we have a horse that will give power 745.7 W
So it will do work of 745.7 J in 1 second of time
now if we wish to find the work done by horse in 0.55 s
so we can say



So it will do total work of 410.14 J in 0.55 s of time
Answer: 3.48g
Explanation:
here, we will be using conservation of momentum to solve the problem. i.e the total momentum remains unchanged, unless an external force acts on the system. We'll in thus question, there is no external force acting in the system.
Remember, momentum = mass * velocity, then
mass of blood * velocity of blood = combined mass of subject and pallet * velocity of subject and pallet
Velocity of blood = 56.5cm = 0.565m
mass of blood * 0.565 = 54kg * (0.000063/0.160)
mass of blood * 0.565 = 54 * 0.00039375
mass of blood * 0.565 = 0.001969
mass of blood = 0.00348kg
Thus, the mass of blood that leaves the heart is 3.48g