Answer:
F = 50636.873 N
Explanation:
given,
bucket of water = 700-kg
length of cable = 20 m
Speed = 40 m/s
angle of the cable = 38.0°
let air resistance be = F
tension in rope be = T
T cos 38° = m×g..................(1)
..........(2)
equation (1)/(2)


F = 50636.873 N
Hence the force exerted on the bucket is equal to F = 50636.873 N
D. a membrane bound nucleus , lmk if im right
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.