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Oksana_A [137]
2 years ago
11

A spring whose stiffness is 980 N/m has a relaxed length of 0. 50 m. If the length of the spring changes from 0. 25 m to 0. 81 m

, what is the change in the potential energy of the spring
Physics
1 answer:
Gelneren [198K]2 years ago
8 0

151.9j

Explanation:

PE=1/2kx^2

PE=1/2(980)(.50)= 245j

PE=(1/2)(980)(.81)= 396.9j

396.9- 245= 151.9j

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13. Determine the kinetic energy of a 2000g roller coaster car that is moving with a speed of 2m/s.
zubka84 [21]

Answer:

4 Joules (if you mean 2000 grams)

4000 Joules (if you mean 2000 kilograms)

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8 0
3 years ago
Un ladrillo se le imparte una velocidad inicial de 6m/s en su trayectoria hacia abajo. ¿cual sera su velocidad final despues de
marshall27 [118]
Saludos!

Respuesta:

28,64 m/s.

Explicación:

Datos: 

Altura o distancia recorrida: 40 m
Vo: 6 m/s 
Aceleración de la gravedad: 9,81 m/s²

El ejercicio puede ser resuelto facilmente utilizando la siguiente formula, sin embargo es posible realizarlo utilizando formulas diferentes. 

Entonces tenemos que:

Vf ^{2} -Vo ^{2} =2 x g x h

Es importante saber que al estar lanzando el ladrillo hacia abajo, el sentido del movimiento sigue el sentido de la gravedad, es decir es necesario que tomes el valor de la gravedad como positivo (+) y no negativo (-) como normalmente se usa.

Sustituyendo tenemos que:

Vf x^{2} =(6 m/s) ^{2} +(2)x(9,81 m/s ^{2})x(40m) \\ Vf ^{2} =820,8 m^{2} /s ^{2}  \\  \sqrt{Vf ^{2} } = \sqrt{820,8m^{2}/s ^{2}  }  \\ Vf=28,64 m/s

Que tengas un buen día!
6 0
3 years ago
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