The snapshot of light as the cart moves with constant velocity is represented by a graph with uniform displacement at each time interval.
The change in displacement with time is uniform at constant velocity. The displacement of the supplied moving item grows at the same pace.
The beginning velocity equals the ultimate velocity at constant velocity.
v₁ = v₂
The object's acceleration at constant velocity is zero since the velocity change with time is zero.
As a result, we may deduce that the graph with equal displacement at each time interval reflects a snapshot of light as the cart moves at a constant speed.
A moving object's displacement-time graph shows the distance traveled by a moving item as time passes. A vector quantity is displacement. The slope or gradient of this graph represents the velocity of the item. The displacement-time graph, also known as the position-time graph, describes an object's motion. In this graph, the displacement of the moving item is displayed on the y-axis as a dependent variable, while time is shown on the x-axis as an independent variable.
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F(x) = f(0) just solve for x
Answer:
d = 90 ft
Explanation:
As we know that after each bounce it reaches to 4/5 times of initial height
so we can say
so the distance covered is given as
here we know that
h = 10 feet
Answer:
298,220 N
Explanation:
Let the force on car three is T_23-T_34
Since net force= ma
from newton's second law we have
T_23-T_34 = ma
therefore,
T_23-T_34 = 37000×0.62
T_23= 22940+T_34
now, we need to calculate
T_34
Notice that T_34 is accelerating all 12 cars behind 3rd car by at a rate of 0.62 m/s^2
F= ma
So, F= 12×37000×0.62= 22940×12= 275280 N
T_23 =22940+T_34= 22940+ 275280= 298,220 N
therefore, the tension in the coupling between the second and third cars
= 298,220 N
