The function of the current with respect to time is the derivative of the charge function with respect to time t. We can apply chain rule to differentiate it:

$I(t) = Q'(t) = (16cos(\omega t + \pi/4))' \\= -16(\omega t + \pi/4)' sin(\omega t + \pi/4) \\= -16\omega sin(\omega t + \pi/4)$

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3 years ago

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A ball of a mass 0.3 kg is released from rest at a height of 8 m. How fast is it going when it hits the ground? Acceleration due

Vaselesa [24]

In order to solve this problem, there are two equations that you need to know to solve this problem and pretty much all of kinematics. The first is that d=0.5at^2 (d=vertical distance, a=acceleration due to gravity and t=time). The second is vf-vi=at (vf=final velocity, vi=initial velocity, a=acceleration due to gravity, t=time). So to find the time that the ball traveled, isolate the t-variable from the d=0.5at^2. Isolate the t and the equation now becomes $\sqrt{(2d)/a}$ . Solving the equation where d=8 and a=9.8 makes the time $\sqrt{(2*8)/9.8}$ =1.355 seconds. With the second equation, the vi=0 m/s, the vf is unknown, a=9.8 m/s^2 and t=1.355 sec. Substitute all these values into the equation vf-vi=at, this makes it vf-0=9.8(1.355). This means that the vf=13.28 m/s.

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3 years ago

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A line dance that is fast in pace will most likely:

Semmy [17]

Increase your heart rate ! I hope this helps

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3 years ago