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3 years ago

10

Suppose that the distance an aircraft travels along a runway before takeoff is given by Upper D equals (5 divided by 3 )t square

dD=(5/3)t2, where D is measured in meters from the starting point and t is measured in seconds from the time the brakes are released. The aircraft will become airborne when its speed reaches 300300 km divided by hkm/h. How long will it take to become airborne, and what distance will it travel in that time?

1 answer:

Dahasolnce [82]3 years ago

4 0

Answer:

a. Time=25seconds

b.distance=1041.67m

Explanation:

a.The equation for $D$ in terms of m/s is $\frac{250}{3}m/s$ after conversion.

To find when speed reaches 300km/hr=83.33m/s, we find $D\prime$ and solve for $t$

$D=\frac{5}{3}t^2\\D\prime=\frac{5}{3}(2t)=\frac{10}{3}t=\frac{250}{3}\\t=25sec$

b. From a, above we already have our t=25seconds as the time it takes before the plane is airborne.

#To find distance travelled in that time , we substitute for $t=25$ in our distance equation:

$D(25)=\frac{5}{3}(25)^2\\=1041.67m$

Hence the distance of the plane before it gets airborne is 1041.67m

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A

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In order to gain a good understanding of the solution above it is necessary to understand that the concept required to solve the question is energy stored in the parallel plate capacitor.

Initially, take the first case. In that, according to the formula for energy stored in parallel plate capacitor with the dielectric inserted between the two plates, find the energy stored. Then, find the energy stored in the parallel plate capacitor when no dielectric is present. Then, write the equation of energy stored in the capacitor with the dielectric present in the form of the energy stored in the capacitor without the dielectric present. The equation must not be in the form of voltage as battery is removed in this case.

For part B, use the equation of the energy dissipated in the resistor. Write it in the form of the equation for energy stored in the parallel plate capacitor without dielectric in it. The equation must be in the form of voltage as battery is kept connected. Looking at the fundamentals

The energy stored in the parallel plate capacitor with the dielectric is given by,

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In this equation, U is the energy stored in the parallel plate capacitor without dielectric, q is charge, and C is the capacitance.

For part B, the battery is still connected. Thus, the equation $q = CV$ is used to modify the above equation.

Thus, the energy stored in the parallel plate capacitor with the dielectric is given by,

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In this equation, $Uk$ is the energy dissipated in the resistor, V is voltage, k is the dielectric constant, and C is the capacitance.

The equation of the energy stored in the parallel plate capacitor without dielectric is,

$U= \frac{1}{2} \frac{C^ 2 V ^2}{C} \\\\= \frac{1}{2} CV ^2$

In this equation, U is the energy dissipated in the resistor, V is voltage, k is the dielectric constant, and C is the capacitance.

(A)

The equation for energy dissipated in the resistor is,

$U _k = \frac{1}{2} \frac{q ^2}{kC}$

Substitute $U = \frac{1}{2}\frac{{{q^2}}}{C}$ in the equation of ${U_k}$

$U _k = \frac{1}{2} (\frac{1}{k} )\frac{q ^2}{C} \\\\= (\frac{1}{k} ) \frac{q^2}{C}\\\\ U_{k} = \frac{U}{k}$

Note :

If the resistance relates to the capacitor, the energy stored in the capacitor is dissipated through the resistance. Thus, by substituting the equation of U, the expression is found out.

(B)

The equation for energy dissipated in the resistor is

$U_{k} = \frac{1}{2}kCV^2$

Here, V is voltage in the circuit.

Substitute $U =\frac{1}{2} CV^2$ in the equation of ${U_k}$

So,

$U_{k} = \frac{1}{2} kCV^2\\$

$= k(\frac{1}{2} CV^2)$

$U_{k} = kU$

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