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angie249
05.03.2020 •
Mathematics
Cual es el volumen en metros cúbicos y en litros de 300 N de aceite de oliva cuyo peso especifico es de 9016 N/ m3
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Ответ:
SPANISH
Como el peso específico del objeto es Y = 9016 N / m ^ 3, puede ver que será Y = 300 N / V m ^ 3
V (m ^ 3) = 300 N / (9016 N / m ^ 3)
V (m ^ 3) = 0, =. 033 m ^ 3
1 Litro es 0,001 m ^ 3
(.033 m ^ 3) * (1 L / 0,001 m ^ 3) = 35,937 L
ENGLISH
Since the specific weight of the object is Y = 9016 N/ m^3 you can see that it will be Y = 300 N / V m^3
V(m^3) = 300 N / (9016 N/m^3)
V(m^3) = 0,=.033 m^3
1 Litro es 0,001 m^3
(.033 m^3) * (1 L / 0,001 m^3) = 35,937 L
Espero que esto ayude, ahora sabes la respuesta y cómo hacerlo. ¡QUE TENGAS UN DÍA BENDITO Y MARAVILLOSO! :-)
- Cutiepatutie ☺❀❤
Ответ:
1:6:3
Step-by-step explanation:
becuase the score of able is one six times grater than ben's and 3 times greater than cal