All categories

3 years ago

Mga halimbawa ng metapora

Physics

2 answers:

faust18 [17]3 years ago

4 0

Answer:

UMm If i understood ide answer

Explanation:

mart [117]3 years ago

3 0

Ang isang talinghaga ay isang pigura ng pagsasalita na, para sa retorikong epekto, direktang tumutukoy sa isang bagay sa pamamagitan ng pagbanggit ng isa pa. Maaari itong magbigay ng kalinawan o makilala ang mga nakatagong pagkakatulad sa pagitan ng dalawang magkakaibang mga ideya. Ang mga talinghaga ay madalas na ihinahambing sa iba pang mga uri ng matalinhagang wika, tulad ng antithesis, hyperbole, metonymy at simile.

1. Ang kanyang mga salita ay pinutol ng mas malalim kaysa sa isang kutsilyo. Ang mga salita ay hindi natutupad sa matatalim na bagay. Sa talinghagang ito, sinabi ng isang tao ang isang bagay na nakasakit sa iba pa.

2. Nararamdaman kong dumarating ang baho ng kabiguan. Ang pagkabigo ay hindi masaya ngunit hindi ito amoy. Kaya, kapag ginamit ang talinghagang ito, nangangahulugan itong malapit na ang isa sa mga pagkabigo ng buhay.

3. Nalulunod ako sa dagat ng kalungkutan. Dito, napakalaki ng kalungkutan na nararamdaman ng tao na walang magawa, tulad ng hinihila siya sa ilalim ng tubig.

4. Dumadaan siya sa isang rollercoaster ng emosyon. Ang aming emosyon ay hindi maaaring sumakay sa isang rollercoaster. Ang talinghagang ito ay nangangahulugan lamang ng pagdaan ng isang tao sa maraming iba't ibang mga kalooban.

You might be interested in

The period T of a simple pendulum depends on the length L of the pendulum and the acceleration of gravity g (dimensions L/P). (a

andriy [413]

Answer:

a). L= meters, g= $\frac{m}{s^{2} }$

b). L= 5cm T=0.448

L= 10 m T=6.34

c). Constant= $\frac{2*\pi }{\sqrt{g} }=\frac{2*\pi }{\sqrt{9.8} }=2.007089923$

Explanation:

a).

$T= 2*\pi \sqrt{\frac{L}{g} } = 2*\pi \sqrt{\frac{m}{\frac{m}{s^{2} } } } \\T= 2*\pi \sqrt{\frac{s^{2}*m }{m} }=2*\pi \sqrt{s^{2} } \\T= s$

b).

$L_{1}= 5 cm$ , $5cm *\frac{1m}{100 cm} = 0.05 m$

$T=2*\pi \sqrt{\frac{L}{g} }$

$T=2*\pi \sqrt{\frac{0.05}{9.8} }= 0.448$ s

$L_{1}= 10 m$

$T=2*\pi \sqrt{\frac{L}{g} }$

$T=2*\pi \sqrt{\frac{10}{9.8} }= 6.43$ s

c).

$g= 9.8 \frac{m}{s^{2} }$

$T=2*\pi *\frac{\sqrt{L} }{\sqrt{g} } =T=2*\pi *\frac{\sqrt{L} }{\sqrt{9.8} } \\T= 2*\pi \frac{1}{\sqrt{9.8}} *\sqrt{L}\\T= 2.007089923*\sqrt{L}$

6 0

2 years ago

I GIVE CROWNS!!<br> I need how you worked it out as well please because I don't know how to do it.

RoseWind [281]

Answer:

hey

Explanation:

5 0

2 years ago

If the velocity of a body changes from 13 m/s to 30 m/s while undergoing constant acceleration, what's the average velocity of t

Ronch [10]

Since the acceleration is constant, the average velocity is simply the average of the initial and final velocities of the body:

$v_{avg} = \frac{v_f+v_i}{2}=\frac{30 m/s+13 m/s}{2}=21.5 m/s$

We can proof that the distance covered by the body moving at constant average velocity $v_{avg}$ is equal to the distance covered by the body moving at constant acceleration a:

- body moving at constant velocity $v_{avg}$ : distance is given by

$S=v_{avg}t = \frac{v_f+v_i}{2}t$

- body moving at constant acceleration $a=\frac{v_f-v_i}{t}$ : distance is given by

$S=v_i t+ \frac{1}{2}at^2 = v_i t + \frac{1}{2}\frac{v_f-v_i}{t}t^2=(v_i+\frac{1}{2}(v_f-v_i))t=\frac{v_f+v_i}{2}t$

7 0

3 years ago

Read 2 more answers

Zorn and Porsha are ice skating. Porsha has a mass of 60 kg, and Zorn has a mass of 40 kg. As they face each

algol13

Assuming the friction between the skaters and the ice is negligible, the magnitude of Porsha's acceleration is 2.8m/s².

Missing part of the question: determine the magnitude of Porsha's acceleration.

Given the data in the question;

Mass of Porsha; $m_{porsha} = 60kg$

Mass of Zorn; $m_{zorn} = 40kg$

Force of Porsha push; $F_{porsha} = 168N$

Magnitude of Porsha's acceleration; $a = \ ?$

To determine the magnitude of Porsha's acceleration, we use Newton's second laws of motion:

$F = m*a$

Where m is the mass of the object and a is the acceleration.

We substitute the mass of Porsha and the force he used into the equation

$168N = 60kg * a\\\\a = \frac{168kg.m/s^2}{60kg}\\\\a = 2.8m/s^2$

Therefore, assuming the friction between the skaters and the ice is negligible, the magnitude of Porsha's acceleration is 2.8m/s².

Learn more: brainly.com/question/25125444

3 0

3 years ago

What type of heat transfer is boiling water??

beks73 [17]

Heat can be transferred from one place to another by three methods:

conduction in solids,

convection of fluids (liquids or gases),

radiation through anything that will allow radiation to pass.

6 0

3 years ago

Read 2 more answers

Mga halimbawa ng metapora​

Mga halimbawa ng metapora