Students are given the following information about a 2.0 kg, motorized toy boat on water. what net force is exerted on the boat
from 4S to 8S?
1 answer:
The motorized toy boat experiences a net force of 0 N between 4 s and 8 s.
The motorized toy boat moves at 8 m/s (u) at 4 s and at 8 m/s (v) at 8 s. We can calculate the acceleration (a) in that period using the following kinematic expression.
The object with a mass (m) of 2.0 kg experiences an acceleration of 0 m/s². We can calculate the net force (F) in that period using Newton's second law of motion.
The motorized toy boat experiences a net force of 0 N between 4 s and 8 s.
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Answer:
y = 67.6 feet, y = 114.4/ (22 - 3t)
Explanation:
For this exercise let's use that light travels in a straight line and some trigonometric relationships, the symbols are in the attached diagram
Large triangle Projector up to the screen
tan θ = y / L
For the small triangle. Projector up to the person
tan θ = y₀ / (L-d)
The angle is the same, so we equate the two equations
y₀ / (L -d) = y / L
y = y₀ L / (L-d)
The distance from the screen (d), we look for it with kinematics
v = d / t
d = v t
we replace
y = y₀ L / (L - v t)
y = 5.2 22 / (22 - 3 t)
y = 114.4 (22 - 3t)⁻¹
This is the equation of the shadow height change as a function of time
For the suggested distance the shadow has a height of
y = 114.4 / (22-13)
y = 67.6 feet
