The total amount of charge in coulombs that has entered a wire at time t is given by the expression Q=4t−t 2 , where t is in seconds and t≥0. Graph I versus t for the interval 0≤t≤4 s.

Hey, everyone. So this problem is dealing with electric current. Let's see what it's asking us. The amount of charge flowing through a conductor over time is given by the function Q of T is equal to five T squared minus three T where T is in seconds and Q cannot exceed a maximum value of 10 columns. We are asked to pot gra of the current versus time over the integral T in which the amount of Q is allowed. So we can look at what this problem gives us. We know the charge as a function is Q sub T is equal to five T squared minus three T. We know the maximum queue, the maximum charge is going to be 10 Kums. And we can recall that our current is equal to the change in Q over or divided by the change in time. And so we can utilize these first two pieces of information to find the interval of time in which the charge reaches its maximum. And so to do that, we can look at Q of T is equal to 10 C which columns, which is equal to five T squared minus three T and then putting that into our quadratic formula format, we have zero equals five T squared minus three T minus 10. And we can use our quadratic formula to factor that we get T is equal to plus or minus 1.7 seconds. So we will take that positive time. We're dealing in positive time. Um No sci fi here and our uh time is 1.7 seconds. Now, we need to figure out what the graph looks like. So we have, or we, we know that our um current as a function of time is equal to the change in charge divided by the change in time. And so we can rewrite that as DDT multiplied or of five T squared minus three T. And then when we take that derivative, we get 10 T minus three. So we have a linear graph with an X where, where it crosses the X axis at 0.3. And so if we have our current on the Y axis and our time on the X axis plot, some, let's put some hash marks here as the time is in second is in cool ones, 5, 1015 and then one and two on the X axis. So our X intercept is at 0.3 and we know between 0.3 and all that 1.7 right there, we can reach our maximum of 10 columns. And so the graph should look something like that. It's going to be linear in fashion. Um And so that's the correct answer for this problem.

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29. Sources of Magnetic Field

Magnetic Field Produced by Moving Charges

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29. Sources of Magnetic Field

Magnetic Field Produced by Moving Charges

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Problem 58b

Textbook Question

The total amount of charge in coulombs that has entered a wire at time t is given by the expression Q=4t−t², where t is in seconds and t≥0. Graph I versus t for the interval 0≤t≤4 s.

Verified step by step guidance

Understand the problem: The charge entering the wire is given by the equation Q = 4t - t^2, where Q is the charge in coulombs and t is the time in seconds. To graph the current (I) versus time (t), recall that current is the rate of change of charge with respect to time, i.e., I = dQ/dt.

Differentiate the given equation Q = 4t - t^2 with respect to time t to find the expression for current I. Using the power rule of differentiation, dQ/dt = d(4t)/dt - d(t^2)/dt. This simplifies to I = 4 - 2t.

Determine the interval of interest for the graph. The problem specifies the interval 0 ≤ t ≤ 4 seconds. Within this range, calculate the values of I at key points (e.g., t = 0, t = 2, t = 4) to understand the behavior of the current.

Analyze the behavior of the current. From the expression I = 4 - 2t, observe that the current decreases linearly with time. At t = 0, I = 4 A; at t = 2, I = 0 A; and at t = 4, I = -4 A. This indicates that the current changes direction after t = 2 seconds.

Plot the graph of I versus t. On the x-axis, represent time (t) in seconds, and on the y-axis, represent current (I) in amperes. Plot the points (0, 4), (2, 0), and (4, -4), and draw a straight line connecting them to represent the linear relationship I = 4 - 2t.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Charge (Q)

Charge is a fundamental property of matter that causes it to experience a force when placed in an electromagnetic field. In this context, the total charge Q is expressed as a function of time t, indicating how charge accumulates in the wire over time. The equation Q=4t−t² shows that charge can increase and then decrease, reflecting the dynamics of charge flow.

Quadratic Functions

A quadratic function is a polynomial function of degree two, typically represented in the form f(t) = at² + bt + c. The expression Q=4t−t² is a quadratic function where the coefficient of t² is negative, indicating that the graph will be a downward-opening parabola. Understanding the properties of quadratic functions, such as their vertex and intercepts, is essential for graphing the charge over time.

Graphing Functions

Graphing functions involves plotting points on a coordinate system to visualize the relationship between variables. For the given function Q=4t−t², one must calculate values of Q for various t within the specified interval (0≤t≤4 s) and then plot these points to create a graph. This visual representation helps in understanding how charge changes over time and identifying key features like maximum charge and when it occurs.

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The total amount of charge in coulombs that has entered a wire at time t is given by the expression Q=4t−t², where t is in seconds and t≥0. Find an expression for the current in the wire at time .

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The total amount of charge in coulombs that has entered a wire at time t is given by the expression Q=4t−t2, where t is in seconds and t≥0. Graph I versus t for the interval 0≤t≤4 s.

Key Concepts

Charge (Q)

Quadratic Functions

Graphing Functions

Watch next

The total amount of charge in coulombs that has entered a wire at time t is given by the expression Q=4t−t², where t is in seconds and t≥0. Graph I versus t for the interval 0≤t≤4 s.