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olga55 [171]
2 years ago
12

All of the dimensions on an aircraft drawing are_________ to the bottom of the drawing.

Engineering
1 answer:
Jet001 [13]2 years ago
7 0
All of the dimensions on an aircraft drawing are _________ to the bottom of the drawing


Answer: parallel
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A mass of 5 kg of saturated liquid-vapor mixture of water is contained in a piston-cylinder device at 125 kPa. Initially, 2 kg o
Likurg_2 [28]

Answer:

See attached pictures.

Explanation:

See attached pictures for detailed explanation.

8 0
4 years ago
The symmetrical load below is connected to a three-phase network. A line current of 25A has been measured. The load resistors ha
boyakko [2]

Answer:

The line voltage of the three phase network is 346.41 V

Explanation:

Star Connected Load

Resistance, R₁ = R₂ = R₃ = 18 Ω

For a star connected load, the line current = the phase current, that is we have

I_L = 25 \, A =  I_{Ph}

Whereby the the voltage across each resistance = V_R is given by the relation;

V_R = I_{Ph} × R

Hence;

V_{Ph} = V_R = I_{Ph} × R  = 25 × 8 = 200 V

Therefore we have;

The line voltage, V_{L} = √3 × V_{Ph} = √3 × 200 = 346.41 V.

Hence, the line voltage of the three phase network = 346.41 V.

3 0
4 years ago
If you are driving down to the sleep downgrade and you have reached the speed of 40 mph , you would apply the setrvice break unt
bezimeni [28]

Answer:

35 miles

Explanation:

When you are driving a truck that has an air brake system you have to keep in mind that when driving down a steep downgrade, the truck will automatically accelerate due to the inclination of the road, so in order to keep the speed to a controllable situation, you need to activate the service brake until you've reached the 35 miles per hour mark.

4 0
3 years ago
What are the factors that influence the amount of elastic energy in an object? Select all
Arte-miy333 [17]

Answer:the answer is A B and C

A B and C

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7 0
3 years ago
An electronic toy is powered by three 1.58-V alkaline cells, each with an internal resistance of 0.0205 Ω, and a 1.53-V carbon-z
mina [271]

Answer:

(a) The current in amperes that flows through the toy's circuit is 0.923A

(b) The power supplied to the toy is 5.78721W

(c) The internal resistance r2 of the failed dry cell is 72Ω

Explanation:

From the circuit diagram attached. We have the electric component:

B1 = 3* 1.58 = 4.74V

B2 = 1.53V

r1 = 3*0.0205 = 0.0615Ω

r2 = 0.105Ω

R = 6.625Ω

Since the internal resistances and the resistor R are connected in series, we can calculate the total resistance RT as

RT = r1 + r2 +R = 0.0615 + 0.105 + 6.625

= 6.7915Ω

Total Voltage supplied to the circuit by both batteries V = B1 + B2 = 4.74 + 1.53 = 6.27V

(a) CIRCUIT CURRENT

The current I, flowing through the circuit is  i =\frac{V}{R_{T}} = \frac{6.27}{6.7915} =0.923A

The current in amperes that flows through the toy's circuit is 0.923A

(b) THE POWER SUPPLIED TO THE TOY

Power P = I*V =0.923*6.27 = 5.78721W

The power supplied to the toy is 5.78721W

(c) THE VALUE OF r2

Due to dry cell failure, the power supplied to the toy is reduced to 0.5W

Now Power P = \frac{V^{2} }{R} . To calculate the new total resistance of the circuit we will make R the subject of the formular

R=\frac{V^{2} }{P} = \frac{6.27^{2} }{0.5} = \frac{39.3129}{0.5} = 78.6258Ω

Remember that RT = r1 + r2 + R

r1 =RT- (R +r2)

r1 = 78.6258 - 6.6865 =71.9393Ω

The internal resistance r2 of the failed dry cell is 72Ω

6 0
3 years ago
Read 2 more answers
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