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Romashka [77]
2 years ago
12

read the excerpt below and answer the question. "no roving foot shall crush thee here, no busy hand provoke a tear." what type o

f figurative language does freneau employ in these lines from “the wild honeysuckle”? a.metaphor b.onomatopoeia c.paradox d.personification
Physics
1 answer:
Aneli [31]2 years ago
3 0
The type of figurative language that Freneau employ in these lines from "The Wild Honeysuckle" is personification. The correct answer would be option D. Why is it personification? Personification is a figure of speech that uses human attributes to something that is not living. In this line, the author attributes the words "roving" and "busy" to foot and hand, respectively. 
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What force must be exerted on the master cylinder of a hydraulic lift to support the weight of a 2000-kg car (a large car) resti
Nataliya [291]

Archimedes principle states that

 

F1 / A1 = F2 / A2

F2 = (A2 / A1) * F1

 

Also, formula for the force is F = mg. Formula for the area of the cylinder is A = πr^2, therefore we get

 

F2 = (πr2^2 / πr1^2) * mg

 

Since the diameter of the cylinders are 2 cm and 24 cm, r1 = 12 and r2 = 1.

 

Substituting the values to the derived equation, we get

 

F2 = (π 1^2 / π 12^2) * 2400 * 9.8

F2 = 163.3333 N

 

 

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6 0
2 years ago
A 817 kg car has four 8.91 kg wheels. When the car is moving, what fraction of the total kinetic energy of the car is due to rot
KATRIN_1 [288]

Answer:

0.0107

Explanation:

We know that

The rotational kinetic energy due to four wheel is

1/2ဃ²I x 4

So

1/4mR²(v/R)² = mv²

But kinetic energy along straight path of the car is 1/2mv²

=> 1/2( 817)v ²

Kc= 408.5v²

So The fraction of total kinetic energy that is due to rotation of the wheel about their axis

Is Kw/Kw+Kc

and Kw = 1/2* 8.91v²= 4.45v²

So 4.45v²/ 4.45v²+ 408.5v²

= 0.0107 as fraction of total kinetic energy

8 0
2 years ago
You may have noticed runaway truck lanes while driving in the mountains. These gravel-filled lanes are designed to stop trucks t
kati45 [8]

Answer:

The  coefficient of kinetic friction  \mu_k =  0.724

Explanation:

From the question we are told that

   The  length of the lane is  l =  36.0 \  m

    The speed of the truck is  v  =  22.6\  m/s

     

Generally from the work-energy theorem we have that  

    \Delta KE  =   N  *  \mu_k *  l

Here N  is the normal force acting on the truck which is mathematically represented as

     \Delta KE is the change in kinetic energy which is mathematically represented as

        \Delta KE =  \frac{1}{2} *  m *  v^2

=>     \Delta KE =  0.5  *  m *  22.6^2

=>      \Delta KE =  255.38m

        255.38m =    m *  9.8  *  \mu_k *   36.0

=>     255.38  =    352.8  *  \mu_k

=>   \mu_k =  0.724

 

6 0
2 years ago
Jo, Daniel and Helen are pulling a metal ring. Jo pulls with a force of 100N in one direction and Daniel with a force of 140N in
Yakvenalex [24]

Answer:

she is pulling with 40 N force

Explanation:

The ring does not move means that the forces are equal.

Let's call Jo's force x

We have the quation

140 = x + 100

x = 40

5 0
2 years ago
In the swing carousel amusement park ride, riders sit in chairs that are attached by a chain to a large rotating drum as shown i
irinina [24]

Answer:\theta =44.068^{\circ}

Explanation:

Given

time taken to complete the circle=7.9 s

radius of circle(r)=15 m

velocity of rider is given by =\frac{2\pi r}{t}

v=\frac{2\pi 15}{7.9}=11.93 m/s

Let us suppose T is the tension in the chain and \thetais the angle which chain makes with vertical

Therefore T\sin \theta =\frac{mv^2}{r}-1

T\cos \theta=mg --2

Divide 1 & 2 we get

tan\theta =\frac{v^2}{rg}

tan\theta =0.968

\theta =44.068^{\circ}

8 0
2 years ago
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