Answer:
Intraductal Papillary Mucinous Neoplasm
F=ma
Therefore the net force = 1000kg × 2 metres per second per second
So F=2000 N
By working with percentages, we want to see how many inches is the center of gravity out of the limits. We will find that the CG is 1.45 inches out of limits.
<h3>What are the limits?</h3>
First, we need to find the limits.
We know that the MAC is 58 inches, and the limits are from 26% to 43% MAC.
So if 58 in is the 100%, the 26% and 43% of that are:
- 26% → (26%/100%)*58in = 0.26*58 in = 15.08 in
- 43% → (43%/100%)*58in = 0.43*58 in = 24.94 in.
But we know that the CG is found to be 45.5% MAC, then it measures:
(45.5%/100%)*58in = 0.455*58in = 26.39 in
We need to compare it with the largest limit, so we get:
26.39 in - 24.94 in = 1.45 in
This means that the CG is 1.45 inches out of limits.
If you want to learn more about percentages, you can read:
brainly.com/question/14345924
Answer:
k = 104.46 N/m
Explanation:
Here we can use energy conservation
so we will have
initial gravitational potential energy = final total spring potential energy
as we know that she falls a total distance of 31 m
while the unstretched length of the string is 12 m
so the extension in the string is given as
so we have
