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natka813 [3]
3 years ago
13

An engineer wishes to design a curved exit ramp for a toll road in such a way that a car will not have to rely on friction to ro

und the curve without skidding. Suppose that a typical car rounds the curve with a speed of 11.7m/s and that the radius of the curve is 50.0m. At what angle should the curve be banked?

Physics
1 answer:
Dafna11 [192]3 years ago
7 0

To solve this problem we will make a graph that allows us to understand the components acting on the body. In this way we will have the centripetal Force and the Force by gravity generating a total component. If we take both forces and get the trigonometric ratio of the tangent we would have the angle is,

T_x = nsinA = \frac{mv^2}{r}

T_y = ncosA = mg

Dividing both.

tan A = \frac{v^2}{rg}

tan A = \frac{11.7^2}{50*9.8}

A = tan^{-1} (0.279367)

A = 15.608\°

Therefore the angle that should the curve be banked is 15.608°

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Answer:

a) 400.4Joules

b) 262.69Joules

Explanation:

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An inductor in an LC circuit has a maximum current of 2.4 A and a maximum energy of 56 mJ.
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Answer:

The energy stored in the capacitor, when the current in the inductor is 1.2 A, is 41.6 mJ.

Explanation:

In a LC oscillating circuit, the energy is stored in the electric field (between the plates of the capacitor) and in the magnetic field (surrounding the wires of the inductor).

At any time, the sum of both energies can be expressed as follows:

E = 1/2 Q² / C   +  1/2 L I²

In this type of circuit, energy oscillates, which means that it is exchanging between both fields all time.

When the capacitor is completely discharged, all the energy is stored in the magnetic field, and at that time, the current is maximum.

The total energy, when I is maximum, can be written as follows:

E = 1/2 L I² (1)

In our case, when I= 2.4A, E= 56 mJ.

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L = 2 *.56 mJ / (2.4)² A² = 20 mH

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As the total energy must be always the same, i.e., 56 mJ, the energy stored in the capacitor, assuming no losses, must be the difference between the total energy and the one stored in the magnetic field:

Ec = 56 mJ - 14.4 mJ = 41.6 mJ

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