Physics High School

Ball A, moving towards the right, collides with
stationary ball B. Ball A bounces back; B moves
off slowly to the right. Which has the greater
mass, A or B?​

Answers

Answer 1

Answer:

ball b has the greater mass.

Explanation:

b only moved because a bumped into it. if a bounced back, it had less mass and b moved slowly.

Related Questions

PLEASE HELP!

Formulate a conclusion about what happens to the pH of a substance when an
acid and base are mixed.

Answers

Answer:

They gon' hate me regardless, that's why I do what I do (what I do)

See me in person i'm flawless (I'm flawless girl), i might just snatch up your dude (hahahaha)

Explanation:

i just want to be....appreciated

Los muelles de un remolque están calibrados para soportar su peso, cuando se carga el remolque con 2100 kg, la fuerza ejercida es de 20580 N comprime los muelles de un remolque 5,5 cm a)Longitud que desciende el remolque si se carga con 28000 N b)Si ha descendido 4,2 cm la carga

Answers

Answer:

a) El remolque desciende 7.4 cm

b) La carga debe ser de 15715.6 N ó 1603.6 kg

Explanation:

Para los cálculos que involucren muelles, se aplica la Ley de Hooke, la cual relaciona el efecto de una Fuerza y el cambio de longitud que esta ejerce,  en un resorte de elasticidad dada.

Escrito en fórmula:

[tex]F=-k \cdot \Delta L[/tex]

Donde:

F es la fuerza ejercida

k es la constante elástica del muelle

ΔL es la variación de longitud del muelle

El problema indica que al cargar 2100 kg se ejerce una fuerza de 20580 N

Esto se corrobora con la 2da ley de Newton y asumiendo una aceleración de gravedad de 9.8 [tex]\frac{m}{s^{2} }[/tex]

[tex]F_{1} =m \cdot a\\F_{1}=2100kg \cdot 9.8\frac{m}{s^2}\\F_{1}=20580N[/tex]

Esta fuerza comprime o reduce la longitud del muelle en 5.5 cm. Usando estos datos en la Ley de Hooke, podemos obtener la constante elástica k:

[tex]F=-k \cdot \Delta L\\20580N=-k \cdot (-0.055m)\\\\k=\frac{20580N}{0.055m}\\k= 374181\frac{N}{m}[/tex]

Ahora ya tenemos los datos para resolver las preguntas:

a) Longitud que desciende el remolque si se carga con 28000 N

Aplicando directamente la formula de la Ley de Hooke:

[tex]F=-k \cdot \Delta L\\\Delta L=\frac{F}{-k} \\\Delta L= \frac{28000N}{-374181\frac{N}{m}} \\\Delta L=-0.074 m = -7.4cm[/tex]

b) Si ha descendido 4,2 cm la carga

En este caso debemos calcular la fuerza necesaria que haga descender el remolque 4.2cm. Nuevamente utilizando la Ley de Hooke con estos nuevos datos:

[tex]F=-k \cdot \Delta L\\F=-374181\frac{N}{m} \cdot (-0.042m)\\F=15715.6N[/tex]

Si queremos saber la carga en kilogramos:

[tex]F = m \cdot a\\m = \frac{F}{a} \\m = \frac{15715.6N}{9.8\frac{m}{s^2} }\\m= 1603.6 kg[/tex]

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