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Reptile [31]
3 years ago
9

For a short time a rocket travels up and to the left at a constant speed of v = 650 m/s along the parabolic path y=600−35x2m, wh

ere x isin m. The origin of polar coordinate system is the same as the origin of the rectangular coordinate system xy.
Part A

Determine the radial component of velocity of the rocket at the instant when its transverse coordinate θ = 60∘, where θ is measured counterclockwise from the x axis.

Express your answer to three significant figures and include the appropriate units.

Part B

Determine the transverse component of velocity of the rocket at the instant when its transverse coordinate θ = 60∘, where θ is measured counterclockwise from the x axis.

Express your answer to three significant figures and include the appropriate units.

Engineering
1 answer:
julia-pushkina [17]3 years ago
4 0

Answer:

Detailed working is shown

Explanation:

The attached file shows a detailed step by step calculation..

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brilliants [131]

Answer:

V_{p (load)} = 28,3 V - 0,7 V = 27,6 V

V_{p (load)} = 27,6 V\\V_{avg} = 17,57 V

The upper diode conduces in the odd half cycles. The lower diode conduces in the even half cycles.

I_{avg} = \frac{V_{avg}}{R_{load}} = \frac{17,57 V}{1000 Ω} = 17,57 mA

Explanation:

The peak voltage after the 6 to 1 step down is V_{p} = \frac{120}{6} \sqrt{2} =  28,3V. Then, the peak voltage of the rectified output is V_{p} - [tex]V_{avg} = \frac{2V_{p (load)} }{\pi}  = \frac{55,2 V}{\pi } = 17,6 VV_{d}[/tex] and according to the statement, the diodes can be modeled to be V_{d} = 0,7 V. Then, the peak voltage in the load is V_{p (load)} = 28,3 V - 0,7 V = 27,6 V.

The upper diode conduces in the odd half cycles. The lower diode conduces in the even half cycles.

The average output voltage is calculated as:

V_{p (load)} = 27,6 V\\V_{avg} = 17,57 V

The average current in the load is calculated as:

I_{avg} = \frac{V_{avg}}{R_{load}} = \frac{17,57 V}{1000 Ω} = 17,57 mA

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Answer:

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