The range of force exerted at the end of the rope is 285.7 N to 1,000 N.
<h3>Net horizontal force of the cylinder</h3>
The net horizontal force of the cylinder when it is at equilibrium position is determined by applying Newton's second law of motion.
∑F = 0
F - μFn = 0
F - 0.2(5,000) = 0
F - 1,000 = 0
F = 1,000 N
The strength of the applied force increases as the number of turns of the rope increases.
minimum force = total force/number of turns of rope
minimum force = 1,000/3.5
minimum force = 285.7 N
Thus, the range of force exerted at the end of the rope is 285.7 N to 1,000 N.
Learn more about Newton's second law of motion here: brainly.com/question/3999427
Answer:
The answer is zero please Give me Brainly
Explanation:
<span>A capacitor with a very large capacitance is in series with a capacitor
that has a very small capacitance.
The capacitance of the series combination is slightly smaller than the
capacitance of the small capacitor. (choice-C)
The capacitance of a series combination is
1 / (1/A + 1/B + 1/C + 1/D + .....) .
If you wisk, fold, knead, and mash that expression for a while,
you find that for only two capacitors in series, (or 2 resistors or
two inductors in parallel), the combination is
(product of the 2 individuals) / (sum of the individuals) .
In this problem, we have a humongous one and a tiny one.
Let's call them 1000 and 1 .
Then the series combination is
(1000 x 1) / (1000 + 1)
= (1000) / (1001)
= 0.999 000 999 . . .
which is smaller than the smaller individual.
It'll always be that way. </span>
We are given with
distance traveled through vacuum = 1.0 m
refractive index of water = 1.33
refractive index of glass = 1.50
refractive index of diamond = 2.42
distance traveled through water is = 1.0/1.33 = 0.75 m
distance traveled through water is = 1.0/1.50 = 0.67 m
distance traveled through water is = 1.0/2.42 = 0.41 m
