Answer: Here this will help you..
Explanation:
1 kg-m/s to kilogram-force meter/second = 1 kilogram-force meter/second
5 kg-m/s to kilogram-force meter/second = 5 kilogram-force meter/second
10 kg-m/s to kilogram-force meter/second = 10 kilogram-force meter/second
20 kg-m/s to kilogram-force meter/second = 20 kilogram-force meter/second
30 kg-m/s to kilogram-force meter/second = 30 kilogram-force meter/second
40 kg-m/s to kilogram-force meter/second = 40 kilogram-force meter/second
50 kg-m/s to kilogram-force meter/second = 50 kilogram-force meter/second
75 kg-m/s to kilogram-force meter/second = 75 kilogram-force meter/second
100 kg-m/s to kilogram-force meter/second = 100 kilogram-force meter/second
Answer:
F = - k (x-xo) a graph of the weight or applied force against the elongation obtaining a line already proves Hooke's law.
Explanation:
The student wants to prove hooke's law which has the form
F = - k (x-xo)
To do this we hang the spring in a vertical position and mark the equilibrium position on a tape measure, to simplify the calculations we can make this point zero by placing our reference system in this position.
Now for a series of known masses let's get them one by one and measure the spring elongation, building a table of weight vs elongation,
we must be careful when hanging the weights so as not to create oscillations in the spring
we look for the mass of each weight
W = mg
m = W / g
and we write them in a new column, we make a graph of the weight or applied force against the elongation and it should give a straight line; the slope of this line is sought, which is the spring constant.
The fact of obtaining a line already proves Hooke's law.
Increase .... decrease .... presumably it's the "best shape" for a body which has been formed by the gravitational force
Answer:
It should fly 8° to west of south at 430km/h
Explanation:
According to the diagram. X components for both velocities must have the same magnitude in order to get the resultant velocity due south.
Solving for α:
α = 8.03°
