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kirill115 [55]
2 years ago
11

A blinking light of constant period is situated on a lab cart. Which diagram best represents a photograph of light, taken every

2 seconds, as the cart moves with constant velocity?
Physics
1 answer:
Karo-lina-s [1.5K]2 years ago
3 0

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.

Learn more about Uniform displacement here:

brainly.com/question/22102874

#SPJ1

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Explanation:

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3 years ago
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Complete Question

The complete question is shown on the first uploaded image  

Answer:

The uncertainty in inverse frequency is  \Delta  [\frac{1}{w} ]=  \frac{3}{2000} \ s

Explanation:

From the question we are told that

   The value of the proportionality constant is  k  = 5  \frac{Hz }{T}

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   The change in this strength of magnetic field is  \Delta B = 3  \ T

The magnetic field is given as

           B  =  \frac{k}{\frac{1}{w} }

Where w is frequency

The uncertainty or error of the field is given as

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The uncertainty in inverse frequency is given  as

           \Delta  [\frac{1}{w} ]  = \frac{\Delta B}{k [\frac{1}{w^2} ]}

                    \Delta  [\frac{1}{w} ]=  \frac{\Delta B}{k (B)^2 }

substituting values

                  \Delta  [\frac{1}{w} ]=  \frac{3}{5 (20)^2 }

               \Delta  [\frac{1}{w} ]=  \frac{3}{2000} \ s

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3 years ago
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Answer:

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Change in time dt = 0.3 sec

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Induced emf is equal to e=N\frac{d\Phi }{dt}=13\times \frac{0.6}{0.3}=26volt

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5 0
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Answer:

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​

8 0
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