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docker41 [41]
4 years ago
9

Suppose the initial position of an object is zero, the starting velocity is 3 m/s and the final velocity was 10 m/s. The object

moves with constant acceleration. Which part of a velocity vs. time graph can be used to calculate the displacement of the object?

Physics
2 answers:
Sloan [31]4 years ago
5 0

In a vleocity Vs time graph, the change in position of an object is given by

x_{1}-x_{2} =\int\limits^T_t {v} \, dt

where the integral can be taken from the graph as

\int\limits^T_t {v} \, dt=area between velocity curve and time axis from t to T


forsale [732]4 years ago
4 0
The area of the rectangle plus the area of the triangle under the line .-.
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Answer:

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Explanation:

Two small objects each with a net charge of +Q exert a force of magnitude F on each other. If r is the distance between them, then the force is given by :

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Now, if one of the objects with another whose net charge is + 4Q is replaced and also the distance between +Q and +4Q charges is increased 3 times as far apart as they were. New force is given by :

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Dividing equation (1) and (2), we get :

\dfrac{F}{F'}=\dfrac{\dfrac{kQ^2}{r^2}}{\dfrac{4kQ^2}{9r^2}}\\\\\dfrac{F}{F'}=\dfrac{kQ^2}{r^2}\times \dfrac{9r^2}{4kQ^2}\\\\\dfrac{F}{F'}=\dfrac{9}{4}\\\\F'=\dfrac{4F}{9}

Hence, the correct option is (d) i.e. " 4F/9"

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