They are similar because they both have a potential difference, both can be redirected/diverted and <span>both have current. </span>
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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Your answer is c holding a brick doesn't contain movement, but energy to grip on it.
hoped it helped!!!
Answer:
The mass of unknown object is 8.62Kg
Explanation:
To develop this problem it is necessary to apply the equations related to the Drag force and the Force of Gravity.
For the given point, that is, the moment at which the terminal velocity is reached, the two forces equalize, that is,
By definition we know that the Drag force is defined as
Where,
Drag coefficient
Density
A =Cross-sectional Area
V = Velocity
In the other hand we have,
Where,
Mass of sphere
Mass of unknown object
Equating the two equations we have to
Re-arrange for m_2,
Our values are given by,
Replacing in the equation we have,
<em>Therefore the mass of unknown object is 8.62Kg</em>
Answer:
j
Explanation:
x = 4 t^2 - 2 t - 4.5
Position at t = 3 s
x = 4 (3)^2 - 2 (3) - 4.5 = 25.5 m
Velocity at t = 3 s
v = dx / dt = 8 t - 2
v ( t = 3 s) = 8 x 3 - 2 = 22 m/s
Acceleration at t = 3 s
a = dv / dt = 8
a ( t = 3 s ) = 8 m/s^2
When is the velocity = 0
v = 0
8 t - 2 = 0
t = 0.25 second
When is the position = 0
x = 0
4 t^2 - 2 t - 4.5 = 0
t = 1.4 second
