Gibbons, small Asian apes, move by brachiation, swinging below a handhold to move forward to thenext handhold. A 9.3kggibbon has
1 answer:
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Missing figure: http://d2vlcm61l7u1fs.cloudfront.net/media/f5d/f5d9d0bc-e05f-4cd8-9277-da7cdda3aebf/phpJK1JgJ.png
Solution:
We need to find the magnitude of the resultant on both x- and y-axis.
x-axis) The resultant on the x-axis is
in the positive direction.
y-axis) The resultant on the y-axis is
in the positive direction.
Both Fx and Fy are positive, so the resultant is in the first quadrant. We can find the angle and so the direction using
from which we find
Solution:
We need to find the magnitude of the resultant on both x- and y-axis.
x-axis) The resultant on the x-axis is
in the positive direction.
y-axis) The resultant on the y-axis is
in the positive direction.
Both Fx and Fy are positive, so the resultant is in the first quadrant. We can find the angle and so the direction using
from which we find
Answer:
The car stops in 7.78s and does not spare the child.
Explanation:
In order to know if the car stops before the distance to the child, you take into account the following equation:
(1)
vo: initial speed of the car = 45km/h
a: deceleration of the car = 2 m/s^2
t: time
xo: initial distance to the child = 25m
x: final distance to the child = 0m
It is necessary that the solution of the equation (1) for time t are real.
You first convert the initial speed to m/s, then replace the values of the parameters and solve the quadratic polynomial for t:
You take the first value t1 because it has physical meaning.
The solution for t is real, then, the car stops in 7.78s and does not spare the child.