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Dominik [7]
3 years ago
9

A jet fighter pilot wishes to accelerate from rest at 5 ggg to reach Mach-3 (three times the speed of sound) as quickly as possi

ble. Experimental tests reveal that he will black out if this acceleration lasts for more than 5.2 ss . Use 331 m/sm/s for the speed of sound
Physics
1 answer:
ad-work [718]3 years ago
7 0

Answer

given,

acceleration, a = 5 g =  5 x 9.8 = 49 m/s²

speed of sound, v = 331 m/s

speed , v = 3 Mach = 3 x 331 m/s = 993 m/s

time, t = 5.2 s

a) Calculating the time period of black out.

  time required by the aircraft to black out

 using equation of motion

  v = u + a t

 initial velocity = 0 m/s

  993 = 0 + 49 t

   t = 20.26 s

the required time to reach 3 Mach speed is more than the given time hence, pilot will black out.

b) now, calculating the maximum speed it can reach in 5.2 s

 using equation of motion

  v = u + a t

 initial velocity = 0 m/s

  v = 0 + 49 x 5.2

 v = 254.8 m/s

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V = IR

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\frac{1}{R_{T}} =  \frac{1}{R_{1}}+ \frac{1}{R_{2}}+ \frac{1}{R_{3}}...+ \frac{1}{R_{n}}

You have three resistors with the following resistance:
65Ω, 25Ω and 170Ω
\frac{1}{R_{T}} = \frac{1}{R_{1}}+ \frac{1}{R_{2}}+ \frac{1}{R_{3}}...+ \frac{1}{R_{n}}

\frac{1}{R_{T}} = \frac{1}{R_{65}}+ \frac{1}{R_{25}}+ \frac{1}{R_{170}}


\frac{1}{R_{T}} =0.0153+0.04+0.006+0.0059
\frac{1}{R_{T}} =0.0613

Get the reciprocal of both sides and divide:

R_{T} =  \frac{1}{0.0613} =16.32

The total resistance then is 16.32Ω

Now that you have the total resistance, you can solve for the total voltage:
V = IR
V = (1.8)(16.32)
V = 29.376V

The emf of the battery is 29.376V


B. To find the resistance in each resistor, just apply Ohm's law again. In a parallel circuit, the voltage is the same, but the current that runs through it is different for each resistor. Now just solve for the current of each using the same voltage.

Resistor 1: 65Ω
I= \frac{V}{R}
I= \frac{29.376}{65}
I= 0.45A

The current flowing through resistor 1 with a resistance of 65Ω is 0.45A.

Resistor 2: 25Ω
I= \frac{V}{R}
I= \frac{29.376}{25}
I= 1.18A
The current flowing through resistor 2 with a resistance of 25Ω is 1.18A.

Resistor 3: 170Ω
I= \frac{V}{R}
I= \frac{29.376}{170}
I= 0.17A

The current flowing through resistor 3 with a resistance of 170Ω is 0.17A.

If you add up all their current it confirms the given that the total current running through all of them is 1.8A.
4 0
3 years ago
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