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3 years ago

Which is the correct way of dual dimensioning using the position method

Engineering

1 answer:

erastova [34]3 years ago

3 0

Answer:

by placing one of the two measurements below the other or by separating them with a stroke

Explanation:

I had a test on this and it was correct

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List three main commodity areas that drive agriculture

Alborosie

Common agricultural commodities include dairy products, wheat, and coffee. You can invest in and “trade” these products virtually, without running a farm and purchasing and storing the items yourself.

3 0

1 year ago

An AC power generator produces 50 A (rms) at 3600 V. The voltage is stepped up to 100 000 V by an ideal transformer and the ener

RSB [31]

Given:

$I_{rms} = 50 A$

voltage, V = 3600V

step-up voltage, V' = 100000 V

Resistance of line, $R = 100\ohm$

Solution:

To calculate % heat loss in long distance power line:

Power produced by AC generator, P = $50\times 3600$ W

P = 180000 W = 180 kW

At step-up voltage, V = 100000V or 100 kV

current, I = $\frac{P}{V'}$

I = $\frac{1800000}{100000}$

I = 1.8 A

Power line voltage drop is given by:

$V_{drop} = I\times R$

$V_{drop} = 1.8\times 100$

$V_{drop} = 180 V$

Power dissipated in long transmission line $P_{dissipated} = V_{drop}\times I$

Power dissipated in long transmission line $P_{dissipated} = 180\times 1.8$ = 324 W

% Heat loss in power line, $P_{loss} = \frac{P_{dissipated}}{P}\times 100$

% Heat loss in power line, $P_{loss} = \frac{324}{180000}\times 100$

$P_{loss} = 0.18%$

5 0

3 years ago

A motorist is driving his car at 60km/hr when he observes that a traffic light 250m ahead turns red. The traffic light is

Alecsey [184]

Explanation:

Okay soo-

Given-

u = 60 km/hr = 60×1000/3600=50/3 m/s

t = 20 s

s = 250 m

a = ?

v = ?

Solution -

Here, acceleration is uniform.

(a) According to 2nd kinematics equation,

s = ut + ½at^2

250 = 50/3 ×20 + 0.5×a×20×20

250-1000/3=200a

(750-1000)/3=200a

a = -250/(3×200)

a = -5/12

a = 0.4167 m/s^2

The required uniform acceleration of the car is 0.4167 m/s^2.

(b) According to 1st kinematics equation

v = u + at

v = 50/3 + (-5/12)×20

v = 50/3-25/3

v = 25/3

v = 8.33 m/s

The speed of the car as it passes the traffic light is 8.33 m/s.

Good luck!

5 0

3 years ago

Which of the following can minimize engine effort in save fuel

pentagon [3]

Answer:

hmm

Explanation:

How to Save Fuel

Keep your vehicle's engine in good condition.

Maintain proper air pressure in your tires.

The faster you travel, the greater your fuel consumption is.

Try to speed up gradually when you stop for a traffic light.

Try to drive smoothly.

Do not allow your engine to idle.

hope this helps

7 0

3 years ago

Read 2 more answers

The boost converter of Fig. 6-8 has parameter Vs 20 V, D 0.6, R 12.5 , L 10 H, C 40 F, and the switching frequency is 200 kHz. (

mr Goodwill [35]

Answer:

a) the output voltage is 50 V

- the average inductor current is 10 A

- the maximum inductor current is 13 A

- the maximum inductor current is 7 A

c) the output voltage ripple is 0.006 or 0.6%V₀

d) the average current in the diode under ideal components is 4 A

Explanation:

Given the data in the question;

a) the output voltage

V₀ = V $_s$ /( 1 - D )

given that; V $_s$ = 20 V, D = 0.6

we substitute

V₀ = 20 / ( 1 - 0.6 )

V₀ = 20 / 0.4

V₀ = 50 V

Therefore, the output voltage is 50 V

- the average inductor current

$I_L$ = V $_s$ / ( 1 - D )²R

given that R = 12.5 Ω, V $_s$ = 20 V, D = 0.6

we substitute

$I_L$ = 20 / (( 1 - 0.6 )² × 12.5)

$I_L$ = 20 / (( 0.4)² × 12.5)

$I_L$ = 20 / ( 0.16 × 12.5 )

$I_L$ = 20 / 2

$I_L$ = 10 A

Therefore, the average inductor current is 10 A

- the maximum inductor current

$I_{Lmax$ = [V $_s$ / ( 1 - D )²R] + [ V

given that, R = 12.5 Ω, V $_s$ = 20 V, D = 0.6, L = 10 μH, T = 1/200 kHz = 5 hz

we substitute

$I_{Lmax$ = [20 / (( 1 - 0.6 )² × 12.5)] + [ (20 × 0.6 × 5) / (2 × 10) ]

$I_{Lmax$ = [20 / 2 ] + [ 60 / 20 ]

$I_{Lmax$ = 10 + 3

$I_{Lmax$ = 13 A

Therefore, the maximum inductor current is 13 A

- The minimum inductor current

$I_{Lmax$ = [V $_s$ / ( 1 - D )²R] - [ V

given that, R = 12.5 Ω, V $_s$ = 20 V, D = 0.6, L = 10 μH, T = 1/200 kHz = 5 hz

we substitute

$I_{Lmin$ = [20 / (( 1 - 0.6 )² × 12.5)] - [ (20 × 0.6 × 5) / (2 × 10) ]

$I_{Lmin$ = [20 / 2 ] -[ 60 / 20 ]

$I_{Lmin$ = 10 - 3

$I_{Lmin$ = 7 A

Therefore, the maximum inductor current is 7 A

c) the output voltage ripple

ΔV₀/V₀ = D/RCf

given that; R = 12.5 Ω, C = 40 μF = 40 × 10⁻⁶ F, D = 0.6, f = 200 Khz = 2 × 10⁵ Hz

we substitute

ΔV₀/V₀ = 0.6 / (12.5 × (40 × 10⁻⁶) × (2 × 10⁵) )

ΔV₀/V₀ = 0.6 / 100

ΔV₀/V₀ = 0.006 or 0.6%V₀

Therefore, the output voltage ripple is 0.006 or 0.6%V₀

d) the average current in the diode under ideal components;

under ideal components; diode current = output current

hence the diode current will be;

$I_D$ = V₀/R

as V₀ = 50 V and R = 12.5 Ω

we substitute

$I_D$ = 50 / 12.5

$I_D$ = 4 A

Therefore, the average current in the diode under ideal components is 4 A

7 0

3 years ago