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alexdok [17]
3 years ago
9

A ball is thrown straight upward and rises to a maximum height of 18.0 m above its launch point. at what height above its launch

point has the speed of the ball decreased to one-half of its initial value?
Physics
1 answer:
slava [35]3 years ago
5 0

A ball is thrown straight upward and rises to a maximum height of 18.0 m

above its launch point. at what height above its launch point has the

speed of the ball decreased to one-half of its initial value?

-----------------------

kinematic equation

v^2=u^2-2as; v0=,s=18,a=-10m/s/s

u=root 360


(u/2)^2=u^2-2as

=>s=13.5m

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Answer:

Explanation:

Given that

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The voltage across the terminals of a 250nF capacitor is푣푣=�50푉푉, 푡푡≤0(푚푚1푒푒−4000푡푡+푚푚2푡푡푒푒−4000푡푡)푉푉, 푡푡 ≥0The initial current
olga2289 [7]

The first part of the question is not complete and it is;

The voltage across the terminals of a 250 nF capacitor is 50 V, A1e^(-4000t) + (A2)te^(-4000t) V, t0, What is the initial energy stored in the capacitor? Express your answer to three significant figures and include the appropriate units. t

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I = e^(-4000t)[0.95 - 1800t]

Explanation:

A) In capacitors, Energy stored is given as;

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Where C is capacitance and v is voltage.

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U(0) = (1/2)C(vo)²

From the question, C = 250 nF and v = 50V

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A1e^(-4000t) + (A2)te^(-4000t)

So, v(0) = A1e^(0) + A2(0)e^(0)

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Thus;

50 = A1

Now for A2; let's differentiate the equation A1e^(-4000t) + (A2)te^(-4000t) ;

And so;

dv/dt = -4000A1e^(-4000t) + A2[e^(-4000t) - 4000e^(-4000t)

Simplifying this, we obtain;

dv/dt = e^(-4000t)[-4000A1 + A2 - 4000A2]

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Substituting the value of A1 = 50V;

0.4 = 0.001(50) - (250 x 10^(-9)A2)

0.4 = 0.05 - (250 x 10^(-9)A2)

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(0.4 + 0.05)/(250 x 10^(-9))= A2

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