Three identical 6.0-kg cubes are placed on a horizontal frictionless surface in contact with one another. The cubes are lined up
from left to right and a force is applied to the left side of the left cube causing all three cubes to accelerate to the right at 2.0 m/s2 . What is the magnitude of the force exerted on the middle cube by the left cube in this case?
1 answer:
Answer:
24 N
Explanation:
= mass of the cube =
Consider the three cubes together as one.
= mass of the three cubes together =
= acceleration of the combination = 2 ms⁻²
= Force applied on the combination
Using Newton's second law
= Force by the left cube on the middle cube
Consider the forces acting on left cube, from the force diagram, we have
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Answer:
3500N
Explanation:
Given parameters:
Mass of driver = 50kg
Speed = 35m/s
Time = 0.5s
Unknown:
Average force the seat belt exerts on her = ?
Solution:
The average force the seat belt exerts on her can be deduced from Newton's second law of motion.
F = mass x acceleration
So;
F = mass x
F = 50 x = 3500N
Answer:
a) > x<-c(1,2,3,4,5)
> y<-c(1.9,3.5,3.7,5.1,6)
> linearmodel<-lm(y~x)
And the output is given by:
> linearmodel
Call:
lm(formula = y ~ x)
Coefficients:
(Intercept) x
1.10 0.98
b)
And if we compare this with the general model
We see that the slope is m= 0.98 and the intercept b = 1.10
Explanation:
Part a
For this case we have the following data:
x: 1,2,3,4,5
y: 1.9,3.5,3.7,5.1, 6
For this case we can use the following R code:
> x<-c(1,2,3,4,5)
> y<-c(1.9,3.5,3.7,5.1,6)
> linearmodel<-lm(y~x)
And the output is given by:
> linearmodel
Call:
lm(formula = y ~ x)
Coefficients:
(Intercept) x
1.10 0.98
Part b
For this case we have the following trend equation given:
And if we compare this with the general model
We see that the slope is m= 0.98 and the intercept b = 1.10