A factory robot drops a 10 kg computer onto a conveyor belt running at 3.1 m/s. The materials are such that
1 answer:
Answer:
Option (c) is the correct choice, as it indicates the object will cover 1.6 m
Explanation:
We are informed that the conveyor runs at a velocity of u = 3.1 m/sec
The robot's mass is 10 kg
with a static friction coefficient of 0.5 and a kinetic friction coefficient of 0.3
Acceleration due to gravity is g = 9.8
The acceleration a is calculated as kinetic friction coefficient multiplied by g, that is, a = 0.3×9.8 = 2.94
According to the third motion equation
s = 1.6 m
[[TAG_39]]Thus, option (c) stands as the correct option[[TAG_40]][[TAG_41]][[TAG_42]]You might be interested in
Answer:
The BMX rider lands 5.4 meters horizontally away from the ramp's end.
Explanation:
The BMX position vector is represented as:
r = (x0 + v0 × t × cos α, y0 + v0 × t × sin α + ½ × g × t²)
Where:
r = position at time t
x0 = initial horizontal position
v0 = initial speed
α = angle of jump
y0 = initial vertical height
g = gravitational acceleration (-9.8 m/s², upward positive)
Refer to the diagram for clarity. At the landing time, the vertical coordinate of the position vector is -2.4 m, measured from the ramp's edge as the origin. Using the vertical component equation for y, one can solve for t, then substitute t to find the horizontal distance.
The vertical position equation:
-2.4 m = 0 + 5.9 m/s × t × sin 40° - ½ × 9.8 m/s² × t²
Rearranged:
0 = -4.9 t² + 5.9 t × sin 40° + 2.4
Solving this quadratic yields:
t = 1.2 seconds
Then, calculate horizontal distance:
x = 0 + 5.9 m/s × 1.2 s × cos 40° = 5.4 m
This means the BMX lands 5.4 meters from the ramp's edge.
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