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3 years ago

Convert 580,000 kW to [GW] with correct number of significant digits.

Engineering

1 answer:

Minchanka [31]3 years ago

4 0

Answer:

$x = 0.58\,GW$

Explanation:

The conversion is:

$x = (580000\,kW)\cdot \left(\frac{1\,GW}{1000000\,kW} \right)$

$x = 0.58\,GW$

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A block of ice weighing 20 lb is taken from the freezer where it was stored at -15"F. How many Btu of heat will be required to c

Rus_ich [418]

Answer:

Heat required =7126.58 Btu.

Explanation:

Given that

Mass m=20 lb

We know that

1 lb =0.45 kg

So 20 lb=9 kg

m=9 kg

Ice at -15° F and we have to covert it at 200° F.

First ice will take sensible heat at up to 32 F then it will take latent heat at constant temperature and temperature will remain 32 F.After that it will convert in water and water will take sensible heat and reach at 200 F.

We know that

Specific heat for ice $C_p=2.03\ KJ/kg.K$

Latent heat for ice H=336 KJ/kg

Specific heat for ice $C_p=4.187\ KJ/kg.K$

We know that sensible heat given as

$Q=mC_p\Delta T$

Heat for -15F to 32 F:

$Q=mC_p\Delta T$

$Q=9\times 2.03(32+15) KJ$

Q=858.69 KJ

Heat for 32 Fto 200 F:

$Q=mC_p\Delta T$

$Q=9\times 4.187(200-32) KJ$

Q=6330.74 KJ

Total heat=858.69 + 336 +6330.74 KJ

Total heat=7525.43 KJ

We know that 1 KJ=0.947 Btu

So 7525.43 KJ=7126.58 Btu

So heat required to covert ice into water is 7126.58 Btu.

8 0

3 years ago

Types of technology include:

Anastaziya [24]

The answer is D-all choices

3 0

3 years ago

In the circuit given below, R1 = 17 kΩ, R2 = 74 kΩ, and R3 = 5 MΩ. Calculate the gain 1formula58.mml when the switch is in posit

Elenna [48]

Answer:a

a) Vo/Vi = - 3.4

b) Vo/Vi = - 14.8

c) Vo/Vi = - 1000

Explanation:

R1 = 17kΩ

for ideal op-amp

Va≈Vb=0 so Va=0

(Va - Vi)/5kΩ + (Va -Vo)/17kΩ = 0

sin we know Va≈Vb=0

-Vi/5kΩ + -Vo/17kΩ = 0

Vo/Vi = - 17k/5k

Vo/Vi = -3.4

║Vo/Vi ║ = 3.4 ( negative sign phase inversion)

R2 = 74kΩ

for ideal op-amp

Va≈Vb=0 so Va=0

(Va-Vi)/5kΩ + (Va-Vo)74kΩ = 0

-Vi/5kΩ + -Vo/74kΩ = 0

Vo/Vi = - 74kΩ/5kΩ

Vo/Vi = - 14.8

║Vo/Vi ║ = 14.8 ( negative sign phase inversion)

Also for ideal op-amp

Va≈Vb=0 so Va=0

Now for position 3 we apply nodal analysis we got at position 1

(Va - Vi)/5kΩ + (Va - Vo)/5000kΩ = 0 ( 5MΩ = 5000kΩ )

-Vi/5kΩ + -Vo/5000kΩ = 0

Vo/Vi = - 5000kΩ/5kΩ

Vo/Vi = - 1000

║Vo/Vi ║ = 1000 ( negative sign phase inversion)

3 0

3 years ago

Find the Rectangular form of the following phasors?

almond37 [142]

Answer:

The angles are missing in the question.

The angles are :

45, 30, 60, 90, -34, -56, 20, -42, -65, -15

P=10, P=5, P=25, P=54, P=65, P=95, P=250, P=8, P=35, P=150

Explanation:

1. P = 10, θ = 45° rectangular coordinates

x = r cosθ , y = r sinθ

So, rectangular form is x + iy

x = P cosθ = 10 cos 45°

= 7.07

y =P sinθ = 10 sin 45°

= 7.07

Therefore, rectangular form

x + iy = 7.07 + i (7.07)

2. P = 5 , θ = 30°

x = 5 cos 30° = 4.33

y = 5 sin 30° = 2.5

So, (x+iy) = 4.33 + i (2.5)

3. P = 25 , θ = 60°

x = 25 cos 60° = 12.5

y = 25 sin 60° = 21.65

So, (x+iy) = 12.5 + i (21.65)

4. P = 54 , θ = 90°

x = 54 cos 90° = 0

y = 54 sin 90° = 54

So, (x+iy) = 0+ i (54)

5. P = 65 , θ = -34°

x = 65 cos (-34°) = 53.88

y = 65 sin (-34°) = -36.34

So, (x+iy) = 53.88 - i (36.34)

6. P = 95 , θ = -56°

x = 95 cos (-56)° = 53.12

y = 95 sin (-56)° = -78.75

So, (x+iy) = 53.12 - i (78.75)

7. P = 250 , θ = 20°

x = 250 cos 20° = 234.92

y = 250 sin 20° = 85.5

So, (x+iy) = 234.92 + i (85.5)

8. P = 8 , θ = (-42)°

x = 8 cos (-42)° = 5.94

y = 8 sin (-42)° = -5.353

So, (x+iy) = 5.94 - i (5.353)

9. P = 35 , θ = (-65)°

x = 35 cos (-65)° = 14.79

y = 35 sin (-65)° = -31.72

So, (x+iy) = 14.79 - i (31.72)

10. P = 150 , θ = (-15)°

x = 150 cos (-15)° = 144.88

y = 150 sin (-15)° = -38.82

So, (x+iy) = 144.88 - i (38.82)

6 0

3 years ago

Explain why the failure of a garden hose occurred near its end and why the tear occurred along its length. Use numerical values

Nataliya [291]

Answer:

Most hydraulic systems develops pressure surges that may surpass settings valve. by exposing the hose surge to pressure above the maximum operating pressure will shorten the hose life.

Explanation:

Solution

Almost all hydraulic systems creates pressure surges that may exceed relief valve settings. exposing the hose surge to pressure above the maximum operating pressure shortens the hose life.

In systems where pressure peaks are severe, select or pick a hose with higher maximum operating pressure or choose a spiral reinforced hose specifically designed for severe pulsing applications.

Generally, hoses are designed or created to accommodate pressure surges and have operating pressures that is equal to 25% of the hose minimum pressure burst.

7 0

3 years ago