Physics Spmnetic!™ 🔭

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This channel belongs to @thespmneticofficial, and a platform for sharing notes and exercises 🤘🏻 For inquiries, directly ask in a discussion group ✨

Red:Sejarah spmnetic!™Malasia1 351 Educación7 479...

📈 Análisis del canal de Telegram Physics Spmnetic!™ 🔭

El canal Physics Spmnetic!™ 🔭 (@physicsspmnotes) en el segmento lingüístico de Malayo es un actor destacado. Actualmente la comunidad reúne a 26 181 suscriptores, ocupando la posición 7 479 en la categoría Educación y el puesto 1 351 en la región Malasia.

📊 Métricas de audiencia y dinámica

Desde su creación el невідомо, el proyecto ha mostrado un crecimiento acelerado, reuniendo a 26 181 suscriptores.

Según los últimos datos del 27 junio, 2026, el canal mantiene una actividad estable. En los últimos 30 días la variación de miembros fue de 360, y en las últimas 24 horas de 28, conservando un alto alcance.

Estado de verificación: No verificado
Tasa de interacción (ER): El promedio de interacción de la audiencia es 8.54%. Durante las primeras 24 horas tras publicar, el contenido suele obtener 5.38% de reacciones respecto al total de suscriptores.
Alcance de las publicaciones: Cada publicación recibe en promedio 2 234 visualizaciones. En el primer día suele acumular 1 408 visualizaciones.
Reacciones e interacción: La audiencia responde de forma activa: el promedio de reacciones por publicación es 5.
Intereses temáticos: El contenido se centra en temas clave como chapter, halo, jawab, physics, soalan.

📝 Descripción y política de contenido

El autor describe el recurso como un espacio para expresar opiniones subjetivas:
“This channel belongs to @thespmneticofficial, and a platform for sharing notes and exercises 🤘🏻 For inquiries, directly ask in a discussion group ✨”

Gracias a la alta frecuencia de actualizaciones (últimos datos recibidos el 28 junio, 2026), el canal mantiene la vigencia y un amplio alcance. La analítica demuestra que la audiencia interactúa activamente con el contenido, lo que lo convierte en un punto de referencia dentro de la categoría Educación.

26 181

Suscriptores

+2824 horas

+1717 días

+36030 días

2 234

Visitas de la publicación

~ 1 40824 horas

~ 1 89448 horas

8.54%

Tasa de compromiso

~ 4

Mensajes por día

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Archivo de publicaciones

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the total resistance must be

Anonymous voting

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the heating wire should have

Anonymous voting

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u should know that the liquid would be responding to acceleration not speed and therefore there is no way to determine which is faster, only which is speeding up, slowing down or maintaining steady speed.

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for those who don't know about history, few years back an admin of physics spmnetic banned a guy who corrected him, and its a shame to this group, and as an spmnetic admin, I apologize from our side...admin here including me can be incorrect, so pls correct us when we are wrong.

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from zianhao, an unknown legend

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Anonymous voting

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Assuming uniform acceleration: Journey OA s = ut + 0.5at² 20 = 0 + 0.5a(1)² a = 40 m/s² v = u + at v = 0 + 40(1) = 40 m/s Journey AB s = ut + 0.5at² 20 = 40(1) + 0.5a(1)² a = -40m/s² v = u + at v = 40 + (-40)(1) = 0 Which means it's accelerating, decelerating and vice versa. If that's the case, then option B is also correct. Unless, we are assuming a non-uniform acceleration, where we don't care how the acceleration changes, but we assume this acceleration brings the velocity at point A to be 20 m/s, then it has constant velocity all the way down the journey, making option A the only answer.

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I'll say there's not enough information

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If you remember this question, I think there's something interesting to discuss. If you recalled, I said that when the car is at A, the velocity cannot be equal to 20 m/s, assuming a constant acceleration along OA. I stopped there and didn't continue with the discussion. However, if the car continues the journey (taking my case to be true), then you'll find out the OA, BC, DE will have acceleration, making B also a correct answer. So, what will be the real case? Turns out, you'll need to have a non-uniform acceleration along journey OA, then you can have the car travelling at 20m/s at A, and later having a constant velocity throughout the journey, making option A the only correct answer. But still, even it's possible to have the car travelling at 20m/s at A, the working did by Harraz in his Physics Spmnetic was still wrong. He didnt understand how to use v = d/t and a = (v-u)/t correctly

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I see a big misconception here. Not only you, many are doing it as well If you want to use v = displacement/ time, you are assuming that the velocity is constant and there is no acceleration But then you find the acceleration for OA, taking v = 20 m/s. Do you guys see the contradiction here? We should apply suvat formulas Journey OA s = 20 u = 0 v = ? a = ? t = 1 s = ut + ½at² 20 = 0 + ½a a = 40 m/s² Assume that the acceleration is constant Then use v = u + at v = 0 + 40(1) = 40 m/s

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Quiz time🔥

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For truck A, if it decelerates from 200 km per hour to 180km, the water will go to the right side , so we cant really eliminate truck A, for other trucks many people gave reasoning on why it could be the fastest

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And also give reasoning on why U can't use any formula here.

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But at least describe how it moves properly

26 176

Ok U actually can't use. Any formula here

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Repost from ꒰ 📚 ˗ˏˋ 𝐉𝐎𝐌 𝐒𝐓𝐔𝐃𝐘 ´ˎ˗ °◞♡ ꒱

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