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scoundrel [369]
3 years ago
11

Three boys of equal strength try to break a rope (and fail) by tying one end to a fence post and tugging on the other end. Three

additional boys (about the same strength as the others) come along and offer their assistance. Which of the following will maximize the tension in the rope?
A. It does not matter: the rope sustains the same tension in both cases.
B. Untie the rope from the post and have three boys tug on one end and the remaining three on the other (like a tug- of-war)
C. Keep one end of the rope tied to the post and have all six boys tug on the other end.
Physics
1 answer:
podryga [215]3 years ago
4 0

The best choice would be C. Keep one end of the rope tied to the post and have all six boys tug on the other end.

Reason:

A is not correct since in the first case which is B, untying the rope from the post and have 3 boys on each end of it tug, will provide equal or a near equal amount of tension since the boys are all within the same strength, however there can be variables that will affect the tension.

That means B would be true according to A, and then for C, having all six boys pull on the other end of the rope would provide greater tension, since if 3 boys of the same strength range can't break the rope when its tied to the fence, we add the other 3 boys of the same strength.

However, if 3 boys are on each end of the rope and of the same strength then the rope will not break since there is an equal amount of net force exerted on the rope.

We know that the fence can withstand the strength of 3 boys but if we add the other 3 boys then it could provide us with a different outcome.

3B < F

3B(2) ≥ F

So, therefore, our best choice would be C since A and B is incorrect.

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Suppose an automobile has 2000-joules of kinetic energy. when it moves at twice the speed, what will be its kinetic energy? what
NeX [460]

Answer:

K.Eₓ = 4 K.E

K.Eₓ = 9 K.E

Explanation:

Th formula for the kinetic energy of a body is given as follows:

K.E = \frac{1}{2}mv^2\\   ---------------equation (1)

where,

K.E = Kinetic Energy of Automobile

m = mass of automobile

v = speed of automobile

For twice speed:

vₓ = 2v

then,

K.E_{x} = \frac{1}{2}mv_{x}^2\\K.E_{x} = \frac{1}{2}m(2v)^2\\K.E_{x} = 4\frac{1}{2}mv^2\\

using equation (1):

<u>K.Eₓ = 4 K.E</u>

For thrice speed:

vₓ = 3v

then,

K.E_{x} = \frac{1}{2}mv_{x}^2\\K.E_{x} = \frac{1}{2}m(3v)^2\\K.E_{x} = 9\frac{1}{2}mv^2\\

using equation (1):

<u>K.Eₓ = 9 K.E</u>

6 0
3 years ago
What happens when two waves meet while they travel through the same medium
Yuki888 [10]

Wave Interference or Interference of wave

7 0
3 years ago
(III) A baseball is seen to pass upward by a window with a vertical speed of If the ball was thrown by a person 18 m below on th
Ghella [55]

Answer:

<em><u>Assuming that the vertical speed of the ball is 14 m/s</u></em> we found the given values:

a) V₀ = 23.4 m/s

b) h = 27.9 m

c) t = 0.96 s

d) t = 4.8 s

 

Explanation:

a) <u>Assuming that the vertical speed is 14 m/s</u> (founded in the book) the initial speed of the ball can be calculated as follows:  

V_{f}^{2} = V_{0}^{2} - 2gh

<u>Where:</u>

V_{f}: is the final speed = 14 m/s

V_{0}: is the initial speed =?

g: is the gravity = 9.81 m/s²

h: is the height = 18 m

V_{0} = \sqrt{V_{f}^{2} + 2gh} = \sqrt{(14 m/s)^{2} + 2*9.81 m/s^{2}*18 m} = 23.4 m/s  

b) The maximum height is:

V_{f}^{2} = V_{0}^{2} - 2gh

h = \frac{V_{0}^{2}}{2g} = \frac{(23. 4 m/s)^{2}}{2*9.81 m/s^{2}} = 27.9 m

c) The time can be found using the following equation:

V_{f} = V_{0} - gt

t = \frac{V_{0} - V_{f}}{g} = \frac{23.4 m/s - 14 m/s}{9.81 m/s^{2}} = 0.96 s

d) The flight time is given by:

t_{v} = \frac{2V_{0}}{g} = \frac{2*23.4 m/s}{9.81 m/s^{2}} = 4.8 s

         

I hope it helps you!    

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