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3 years ago

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Nikola Tesla and Thomas Edison were both pioneers when it came providing electricity to the people. Edison had hired the genius

Tesla to work for him, but their differences, pride and personalities caused them to compete and publicly insult one another instead of working together. List 5 strategies these two could have used to help them work with others they disagree with.

1 answer:

Pepsi [2]3 years ago

6 0

<u>Explanation:</u>

In other to avoid a situation where their differences, pride, and personalities caused them to compete and publicly insult one another instead of working together they need to respectfully consider the viewpoints of each other, as well as assign individual roles that should be respected.

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What is equilibrium state

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Answer:

A state of rest or balance due to the equal action of opposing forces.

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2 years ago

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An engineer wishes to design a roller coaster so that

nataly862011 [7]

To prevent cars from falling, the radius at the top of the circle should be

small such cars inverted at the top remain attached during motion.

Correct response;

The radius of the coaster can be <u>C) 20 m</u>

<h3>Method by which the above option is selected</h3>

Mass of roller coaster car, m = 500 kg

Speed at the top of the circle, v = 20 m/s

Required:

The maximum radius of the circular path the roller coaster car.

Solution:

$\displaystyle Centrifugal \ force, \, F_{c} = \mathbf{\frac{m \cdot v^2}{r}}$

Where;

r = The radius of the circular path.

Weight of the roller coaster car = m·g = The centripetal force

Where;

g = Acceleration due to gravity = 9.81 m/s²

At equilibrium, we have;

Centrifugal force = Centripetal force

$\displaystyle \frac{m \cdot v^2}{r} = \mathbf{ m \cdot g}$

Therefore;

$\displaystyle r = \mathbf{ \frac{v^2}{g}}$

Which gives;

$\displaystyle r = \frac{20^2}{9.81} \approx 40.77$

The maximum radius for safety of a roller coaster, r ≈ 40.77 meters

$\displaystyle Range \ of \ radius \ of \ the \ circle = \frac{40.77}{4} \leq Radius \ of \ circle \leq 40.77$

Which gives;

Range of the radius of the circle = 10.2 ≤ Radius of circle ≤ 40.77

The correct option for safety considerations is therefore;

<u>C) 20 m</u>

<em>The possible question options are;</em>

<em>A) 5 m B) 10 m C) 20 m D) 40 m E) 80 m</em>

Learn more about centripetal force here:

brainly.com/question/12674230

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2 years ago

A 220 kg crate hangs from the end of a rope of length L = 14.0 m. You push horizontally on the crate with a varying force F to m

kifflom [539]

<u>Answer</u>:

(a) magnitude of F = 797 N

(b)the total work done W = 0

(c)work done by the gravitational force = -1.55 kJ

(d)the work done by the pull = 0

(e) work your force F does on the crate = 1.55 kJ

<u>Explanation</u>:

<u>Given</u>:

Mass of the crate, m = 220 kg

Length of the rope, L = 14.0m

Distance, d = 4.00m

<u>(a) What is the magnitude of F when the crate is in this final position</u>

Let us first determine vertical angle as follows

=> $Sin \theta = \frac{d }{L}$

=> $\theta = Sin^{-1} \frac{d}{L}$ =

Now substituting thje values

=> $\theta = Sin^{-1} \frac{4}{12}$ =

=> $\theta = Sin^{-1} \frac{1}{3}$

=> $\theta = Sin^{-1}(0.333)$

=> $\theta = 19.5^{\circ}$

Now the tension in the string resolve into components

The vertical component supports the weight

=> $Tcos\theta = mg$

=> $T = \frac{mg}{cos\theta}$

=> $T = \frac{230 \times 9.8 }{cos(19.5)}$

=> $T = \frac{2254 }{cos(19.5)}$

=> $T = \frac{2254 }{0.9426}$

=>T =2391N

Therefore the horizontal force

$F = TSin(19.5)$

F = 797 N

b) The total work done on it

As there is no change in Kinetic energy

The total work done W = 0

<u>c) The work done by the gravitational force on the crate</u>

The work done by gravity

Wg = Fs.d = - mgh

Wg = - mgL ( 1 - Cosθ )

Substituting the values

= $-230 \times 9.8\times 12 ( 1 - cos(19.5) )$

= $-230 \times 9.8\times 12 ( 1 - 0.9426) )$

= $-230 \times 9.8\times 12 (0.0574)$

= $-230 \times 9.8\times 0.6888$

= $-230 \times 6.750$

= -1552.55 J

The work done by gravity = -1.55 kJ

<u>d) the work done by the pull on the crate from the rope</u>

Since the pull is perpendicular to the direction of motion,

The work done = 0

e)Find the work your force F does on the crate.

Work done by the Force on the crate

WF = - Wg

WF = -(-1.55)

WF = 1.55 kJ

<u>(f) Why is the work of your force not equal to the product of the horizontal displacement and the answer to (a)</u>

Here the work done by force is not equal to F*d

and it is equal to product of the cos angle and F*d

So, it is not equal to the product of the horizontal displacement and the answer to (a)

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A soccer player is running upfield at 10 m/s and comes to a stop in 3 seconds facing the same direction. What is his acceleratio

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His acceleration would be 3.34 m/s

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Answer:

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