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4 years ago

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The magnetic flux through a metal ring varies with time t according to ΦB = at3 − bt2, where ΦB is in webers, a = 6.00 Wb s−3, b

= 18.0 Wb s−2, and t is in seconds. The resistance of the ring is 2.80 Ω. For the interval from t = 0 to t = 2.00 s, determine the maximum current induced in the ring.

1 answer:

pychu [463]4 years ago

6 0

To solve this problem it is necessary to apply the second derivative of the function to find the maximum time reference and thus calculate the maximum voltage.

With the maximum voltage by Ohm's Law it is possible to find the maximum current.

Ohm's law defines that

E = I*R

Where,

I = Current

R= Resistance

On the other hand by faraday studies and the potential can be expressed at the rate of change of the electric flow, that is

$E = \frac{d\phi}{dt}$

Replacing with our values we have that

$E = \frac{d(6t^3-18t^2)}{dt}$

$E = -18t^2 +36t$

The second derivative is

$E' = -36t+36$

When E' = 0 we have a Maximum, then

0 = -36t+36

t = 1

Therefore when the time is 1s E has a Maximum, replacing at the function

$E(t) = -18t^2 +36t$

$E(1) = -18(1)^2 +36(1)$

$E = 18V$

Then the maximum current will be given by

$I = \frac{E}{R}$

$I = \frac{18}{2.8}$

$I = 6.42A$

Therefore the maximum current induced in the ring is 6.42A

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A roadrunner at rest suddenly spots a rattlesnake slithering directly away at a constant speed of 0.75 m/s. At the moment the sn

Shalnov [3]

Answer:

The roadrunner will take approximately 5.285 seconds to catch up to the rattlesnake.

Explanation:

From the statement we notice that:

1) Rattlesnake moves a constant speed ( $v_{S} = 0.75\,\frac{m}{s}$ ), whereas the roadrunner accelerates uniformly from rest. ( $v_{o, R} = 0\,\frac{m}{s}$ , $a = 1\,\frac{m}{s^{2}}$ )

2) Initial distance between the roadrunner and rattlesnake is 10 meters. ( $x_{o, R} = 0\,m$ , $x_{o,S} = 10\,m$ )

3) The roadrunner catches up to the snake at the end. ( $x_{S} = x_{R}$ )

Now we construct kinematic expression for each animal:

Rattlesnake

$x_{S} = x_{o,S}+v_{S}\cdot t$

Where:

$x_{o, S}$ - Initial position of the rattlesnake, measured in meters.

$x_{S}$ - Final position of the rattlesnake, measured in meters.

$v_{S}$ - Speed of the rattlesnake, measured in meters per second.

$t$ - Time, measured in seconds.

Roadrunner

$x_{R} = x_{o,R} +v_{o,R}\cdot t +\frac{1}{2}\cdot a\cdot t^{2}$

Where:

$x_{o, R}$ - Initial position of the roadrunner, measured in meters.

$x_{R}$ - Final position of the roadrunner, measured in meters.

$v_{o,R}$ - Initial speed of the roadrunner, measured in meters per second.

$a$ - Acceleration of the roadrunner, measured in meters per square second.

$t$ - Time, measured in seconds.

By eliminating the final positions of both creatures, we get the resulting quadratic function:

$x_{o,S}+v_{S}\cdot t = x_{o,R}+v_{o,R}\cdot t +\frac{1}{2}\cdot a \cdot t^{2}$

$\frac{1}{2}\cdot a \cdot t^{2} + (v_{o,R}-v_{S})\cdot t + (x_{o,R}-x_{o,S}) = 0$

If we know that $a = 1\,\frac{m}{s^{2}}$ , $v_{o, R} = 0\,\frac{m}{s}$ , $v_{S} = 0.75\,\frac{m}{s}$ , $x_{o, R} = 0\,m$ and $x_{o,S} = 10\,m$ , the resulting expression is:

$0.5\cdot t^{2}-0.75\cdot t -10=0$

We can find its root via Quadratic Formula:

$t_{1,2} = \frac{-(-0.75)\pm \sqrt{(-0.75)^{2}-4\cdot (0.5)\cdot (-10)}}{2\cdot (0.5)}$

$t_{1,2} = \frac{3}{4}\pm \frac{\sqrt{329}}{4}$

Roots are $t_{1} \approx 5.285\,s$ and $t_{2}\approx -3.785\,s$ , respectively. Both are valid mathematically, but only the first one is valid physically. Hence, the roadrunner will take approximately 5.285 seconds to catch up to the rattlesnake.

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