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ludmilkaskok [199]

1 year ago

8

In at least 150 words, discuss how Chang's use of personification, metaphor, or connotation express one his themes in "Garden of

My Childhood." Use details from the poem to support your response.

1 answer:

Kay [80]1 year ago

6 0

<span>Poet Kuangchi Chang did not remain in China long enough to be "re-educated." Following the Communist takeover he fled to the United States. His poem "Garden of My Childhood" describes China before the revolution as a peaceful, idyllic garden with a violent horde rapidly approaching. A vine, the wind, and the sea are each personified, and each beckons for him to run. It is not until "eons later," when he is "worlds away," that his "running is all done," and he finds himself at his destination: another garden, just like the one he had left behind.</span>

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A rocket starts from rest and moves upward from the surface of the earth. for the first 10.0 s of its motion, the vertical accel

Vikki [24]

acceleration of rocket is given here as

$a_y = 2.60* t$

now we know that

$\frac{dv}{dt} = 2.60t$

now integrating both sides

$\int dv = \int 2.60t dt$

$v = 2.60\frac{t^2}{2}$

$v = 1.30 t^2$

here since its given that rocket will accelerate for t = 10 s

so here we have

$v = 1.30 * 10^2$

$v = 130 m/s$

so after t = 10 s the speed of rocket will be 130 m/s upwards

5 0

1 year ago

A 1500-kg car locks its brakes and skids to a stop on a slippery horizontal road, leaving skid marks that are 15 m long. How muc

Harman [31]

Answer:

$E=88200\ J$

Explanation:

Given:

mass of car, $m=1500\ kg$

distance of skidding after the application of brakes, $d=15\ m$

coefficient of kinetic friction, $\mu_k=0.4$

<u>So, the energy dissipated during the skidding of car:</u>

<em>Frictional force:</em>

$f=\mu_k.N$

where N = normal reaction by ground on the car

$f=0.4\ties 1500\times 9.8$

$f=5880\ N$

<em>Now from the work-energy equivalence:</em>

$E=f.d$

$E=5880\times 15$

$E=88200\ J$ is the dissipated energy.

3 0

1 year ago

A car of mass 1100kg moves at 24 m/s. What is the braking force needed to bring the car to a halt in 2.0 seconds? N

LenaWriter [7]

13200N

Explanation:

Given parameters:

Mass = 1100kg

Velocity = 24m/s

time = 2s

unknown:

Braking force = ?

Solution:

The braking force is the force needed to stop the car from moving.

Force = ma = $\frac{mv}{t}$

m is the mass of the car

v is the velocity

t is the time taken

Force = $\frac{1100 x 24}{2}$ = 13200N

Learn more:

Force brainly.com/question/4033012

#learnwithBrainly

8 0

2 years ago

A stone falls from rest from the top of a cliff. A second stone is thrown downward from the same height 2.7 s later with an init

Darina [25.2K]

Answer:4.05 s

Explanation:

Given

First stone is drop from cliff and second stone is thrown with a speed of 52.92 m/s after 2.7 s

Both hit the ground at the same time

Let h be the height of cliff and it reaches after time t

$h=\frac{gt^2}{2}$

For second stone

$h=52.92\times \left ( t-2.7\right )+\frac{g\left ( t-2.7\right )^2}{2}$ ---2

Equating 1 &2 we get

$\frac{gt^2}{2}=52.92\times \left ( t-2.7\right )+\frac{g\left ( t-2.7\right )^2}{2}$

$\frac{g}{2}\left ( t-t+2.7\right )\left ( 2t-2.7\right )-\left ( t-2.7\right )52.92=0$

$13.23\times \left ( 2t-2.7\right )-\left ( t-2.7\right )52.92=0$

$26.46t-35.721-52.92t+142.884=0$

$t=4.05 s$

4 0

1 year ago

A machine produces photo detectors in pairs. Tests show that the first photo detector is acceptable with probability 3/5. When t

klasskru [66]

Answer:

a. $a. \ \frac{7}{25}$

$b.\ \ \ P(D_1D_2)=\frac{6}{25}$

Explanation:

a. Find the probability that exactly one photo detector of a pair is acceptable:

Let $A_i=i^{th}$ photo is accepted and the probability $D_i=i^{th}$ is defected.

Therefore:

$P(A_i)=3/5,\ P(A_2|A_1)=4/5,\ \ P(A_2|D_1)=2/5\\\\\\=P(A_1D_2)+P(D_1A_2)\\\\=\frac{3}{5}\times\frac{1}{5}+\frac{2}{5}\times\frac{2}{5}\\\\=\frac{7}{25}$

#The probability of exactly one photo detector of a pair is accepted is 7/25

b.Find the probability that both photo detectors in a pair are defective,P(D1D2):

$P(D_1D_2)=\frac{2}{5}\times \frac{3}{5}\\\\=\frac{6}{25}$

Hence, from out tree diagram,the probability that both photo detectors in a pair are defective is 6/25

4 0

2 years ago