Answer:
15.7m/s
Explanation:
To solve this problem, we use the right motion equation.
Here, we have been given the height through which the ball drops;
Height of drop = 14.5m - 1.9m = 12.6m
The right motion equation is;
V² = U² + 2gh
V is the final velocity
U is the initial velocity = 0
g is the acceleration due to gravity = 9.8m/s²
h is the height
Now insert the parameters and solve;
V² = 0² + 2 x 9.8 x 12.6
V² = 246.96
V = √246.96 = 15.7m/s
A horizontal line on a position-time graph represents a stationary object or, in this case, a parked car. The car is not moving since there's no change in position over time. Speed or velocity would be depicted by the slope of the line on the position-time graph.
Your question pertains to understanding what a horizontal line on a position-time graph indicates about a car's motion. In the context of a position-time graph, a horizontal line suggests that the position of the car is not changing with respect to time, indicating option A: The car is not moving. The car being stationary means there is no change in position over time, hence the horizontal line on the graph.
Options B, C, and D suggest a change in position over time, which would be represented by a slanted line, not a horizontal one. Option B: 'moving at a constant speed' and Option C: 'moving at a constant velocity' would both produce a straight line with a slope, not a horizontal line. Option D: 'speeding up' would be represented by a curved line, showing acceleration.
In contrast, a stationary object or a car at rest is represented by a horizontal line on a position-time graph. Velocity or speed, on the other hand, is determined by the slope of the line on a position-time graph. Hence, a horizontal line, having no slope, signifies zero speed or velocity, indicating that the car is not moving.
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lithospheric theory
plate tectonics theory
nebular theory
Plate tectonics theory
b. The potential energy of the body is 30,000 J.
c. The amount of work is 450,000 J.
d. The power exerted is 19,600 W.
B) fission.
C) fusion.
D) isotope.