An object travels along a straight, horizontal surface with an initial speed of 2 ms. The position of the object as a function o
f time is given in the table. Which of the following graphs represents the object’s velocity as a function of time?
1 answer:
Answer:
The options are not provided, so i will answer in a general way.
We know that:
The movement is along a straight horizontal surface, then we have one-dimensional motion.
The speed is 2m/s
We want a graph of position vs time.
Now, remember the relation:
Distance = Speed*Time
Then we can write the position as a function of time as:
P(t) = 2m/s*t + P0
Where t is our variable, that represents time in seconds, and P0 is the position at time t = 0seconds, we can assume that this is zero.
Then the equation is:
P(t) = 2m/s*t
And the graph is something like:
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Answer:
Explanation:
Given that,
A point charge is placed between two charges
Q1 = 4 μC
Q2 = -1 μC
Distance between the two charges is 1m
We want to find the point when the electric field will be zero.
Electric field can be calculated using
E = kQ/r²
Let the point charge be at a distance x from the first charge Q1, then, it will be at 1 -x from the second charge.
Then, the magnitude of the electric at point x is zero.
E = kQ1 / r² + kQ2 / r²
0 = kQ1 / x² - kQ2 / (1-x)²
kQ1 / x² = kQ2 / (1-x)²
Divide through by k
Q1 / x² = Q2 / (1-x)²
4μ / x² = 1μ / (1 - x)²
Divide through by μ
4 / x² = 1 / (1-x)²
Cross multiply
4(1-x)² = x²
4(1-2x+x²) = x²
4 - 8x + 4x² = x²
4x² - 8x + 4 - x² = 0
3x² - 8x + 4 = 0
Check attachment for solution of quadratic equation
We found that,
x = 2m or x = ⅔m
So, the electric field will be zero if placed ⅔m from point charge A, OR ⅓m from point charge B.
