Un movil de masa 12Kg sobre el cual estan actuando varias fuerzas F_1=48N, F_2=60N y F_3=30N Calcular la aceleracion con la cual
1 answer:
Answer:
Lamentablemente el problema está incompleto, pues no sabemos la dirección en la que se aplican las fuerzas. Por ello, voy a resolver el problema asumiendo dos casos. (abajo se puede ver una imagen donde se describe cada caso)
1) Todas las fuerzas están en la misma dirección.
Entonces la fuerza neta será la suma de las 3 fuerzas, entonces:
F = 48N + 60N + 30N = 138N
Y por la segunda ley de Newton sabemos que:
F = m*a
fuerza igual a masa por aceleración.
Entonces la aceleración está dada por:
a = F/m = 138N/12kg = 11.5 m/s^2
2) Segundo caso, suponemos que F1 es opuesta a F2 y F3
En este caso, la fuerza neta será:
F = F2 + F3 - F1 = 60N + 30N - 48N = 42N
En este caso, la aceleración será:
a = 42N/12kg = 3.5 m/s^2
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Answer:
The 5-number summary is
1. Median = 0.93 W/kg
2. Lower quartile = 0.69 W/kg
3. Upper quartile = 1.16 W/kg
4. Minimum value = 0.54 W/kg
5. Maximum value = 1.42 W/kg
Explanation:
We are given the measured radiation absorption rates (in W/kg) corresponding to 11 cell phones.
1.16 0.85 0.69 0.75 0.95 0.93 1.18 1.17 1.42 0.54 0.57
What is 5-number summary?
A 5-number summary refers to a box plot that basically shows 5 statistical characteristics of a data set.
These statistical characteristics are:
1. Median
2. Lower quartile
3. Upper quartile
4. Minimum value
5. Maximum value
1. Median:
Arrange the data in ascending order
0.54 0.57 0.69 0.75 0.85 0.93 0.95 1.16 1.17 1.18 1.42
(n+1)/2 gives the median value of the data set.
(11 + 1)/2 = 6th position
Therefore, 0.93 W/kg is the median of the data set.
2. Lower quartile:
Divide the data set into two equal halfs (include median in both if n = odd)
Lower half = 0.54 0.57 0.69 0.75 0.85 0.93
Upper half = 0.93 0.95 1.16 1.17 1.18 1.42
The lower quartile is the median of the lower half of the data set.
Lower half = 0.54 0.57 0.69 0.75 0.85 0.93
The median is 6/2 = 3rd position
Therefore, the lower quartile of the data set is 0.69 W/kg
3. Upper quartile:
Divide the data set into two equal halfs (include median in both if n = odd)
Lower half = 0.54 0.57 0.69 0.75 0.85 0.93
Upper half = 0.93 0.95 1.16 1.17 1.18 1.42
The upper quartile is the median of the lower half of the data set.
Upper half = 0.93 0.95 1.16 1.17 1.18 1.42
The median is 6/2 = 3rd position
Therefore, the upper quartile of the data set is 1.16 W/kg
4. Minimum value:
The minimum value is the least value in the data set.
0.54 0.57 0.69 0.75 0.85 0.93 0.95 1.16 1.17 1.18 1.42
Therefore, the minimum value of the data set is 0.54 W/kg
5. Maximum value
The maximum value is the least value in the data set.
0.54 0.57 0.69 0.75 0.85 0.93 0.95 1.16 1.17 1.18 1.42
Therefore, the maximum value of the data set is 1.42 W/kg
The box plot is illustrated in the attached diagram.
