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3 years ago

The force component along the displacement varies with the magnitude of the displacement, as shown in the graph. (a) 0 to 1.0 m,

1 answer:

5 0

Work is force*displacement if the force and displacement is parallel.

a. You can average the force over the distance so W = Fave*d

<span>b The force part of that multiplication is zero. </span>

<span>c. You can form the average force for the interval from 2 to 3 and find the work for that section and then consider the interval from 3 to 4, find the work and add the 2 work results.

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

</span>

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A small pot of water has an initial temperature of 80 ° C. It is placed in a cold room where the temperature is 10 ° C. During t

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c. decreases 45°C

Explanation:

because it just makes sense

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1. Choose two forces and compare and contrast these forces. You must provide two ways that they are alike and two ways that they

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what are the forces ? :l

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3 years ago

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Determine the electrical force of attraction between two balloons with separate charges of +3.5 × 10-8 C and -2.9 × 10-8 C when

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Answer:

The electrical force of attraction between the balloons is $2.16\times 10^{-5}\ N$ .

Explanation:

Given that,

Charge 1, $q_1=+3.5\times 10^{-8}\ C$

Charge 2, $q_2=-2.9\times 10^{-8}\ C$

Distance between the charges, d = 0.65 m

We need to find the electrical force of attraction between two balloons. It is given by the formula as :

$F=k\dfrac{q_1q_2}{d^2}\\\\F=9\times 10^9\times \dfrac{3.5\times 10^{-8}\times 2.9\times 10^{-8}}{(0.65)^2}\\\\F=2.16\times 10^{-5}\ N$

So, the electrical force of attraction between the balloons is $2.16\times 10^{-5}\ N$ . Hence, this is the required solution.

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3 years ago

Electrical wires suspended between two towers form a catenary (see figure) modelled by the equation shown below, where x and y a

RideAnS [48]

Given that the arc length is:
s = ∫ √[1² + (dy/dx)²] dx

<span>So arc length between the two points is then: </span>
<span>s = 2*20sinh(x/20)
= 40sinh(x/20) </span>

<span>The straight distance between the two points is : d = 2x </span>
<span>So, x = d/2.
= 40/2
= 20 m </span>

<span>Plug this into the arc length equation to get:
s = 40sinh[20/20)]
=40* ½ (e - 1/e) </span>
<span> = 47 m</span>

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3 years ago

Suppose that the loudspeaker in the problem had a mass of 500 kg and the ropes hung 20∘ from the vertical. Into which of the fol

nexus9112 [7]

The following intervals would you expect T(tension) to fall : <u>2000 to 4000 N </u>

<h3>Further explanation</h3>

The force acting on a system with static equilibrium is 0

$\large{\boxed{\bold{\sum F=0}}$

(forces acting as translational motion only, not including rotational forces)

$\displaystyle \sum F_x = 0\\\\ \sum F_y = 0$

We complete the questions:

A 20-kg loudspeaker is suspended 2.0 m below the ceiling by two ropes that are each 30? from vertical.

Find the value of T, the magnitude of the tension in either of the ropes.

Express your answer in newtons.

Suppose that the loudspeaker in the problem has a mass of 500 kg and the ropes hung 20? from the vertical. Into which of the following intervals would you expect T to fall? You don't have to calculate Texactly to answer this question; just make an estimate.

500 to 1000 N

1000 to 2000 N

2000 to 4000 N

4000 to 8000 N

8000 to 16,000 N

In a 500-kg loudspeaker system and two ropes that are each 20° from vertical, we see the forces acting on the y axis (vertical)

$\displaystyle \sum F_y = 0\\\\T1_y+T2_y-w=0\\\\T1~cos~20^o+T2~cos~20^o=m\times g$

we assume g = 10 m/s², then :

$\displaystyle 2T~cos~20=500\times 10\\\\T~cos~20=2500\\\\\boxed{\bold{T=26592 N}}}$

So the value of T is between 2000 and 4000 N

<h3>Learn more</h3>

Newton's Law

brainly.com/question/13725525

Keywords : the loudspeaker, ropes, Tension, mass, intervals

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4 years ago