Where do greywater pipes generally feed into?

What is the purpose of the graphic language?

solmaris [256]

Answer:

enables the representation, analysis and communication of various aspects of an information system. These aspects correspond to varying and incomplete views of information systems and the processes therein.

5 0

3 years ago

Knowing that the central portion of the link BD has a uniform cross-sectional area of 800 2, determine the magnitude of the load

IceJOKER [234]

Answer: 50

Explanation:

7 0

3 years ago

Find the time-domain sinusoid for the following phasors:_________

sattari [20]

Answer:

a. r(t) = 6.40 cos (ωt + 38.66°) units

b. r(t) = 6.40 cos (ωt - 38.66°) units

c. r(t) = 6.40 cos (ωt - 38.66°) units

d. r(t) = 6.40 cos (ωt + 38.66°) units

Explanation:

To find the time-domain sinusoid for a phasor, given as a + bj, we follow the following steps:

(i) Convert the phasor to polar form. The polar form is written as;

r∠Ф

Where;

r = magnitude of the phasor = $\sqrt{a^2 + b^2}$

Ф = direction = tan⁻¹ ( $\frac{b}{a}$ )

(ii) Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid (r(t)) as follows:

r(t) = r cos (ωt + Φ)

Where;

ω = angular frequency of the sinusoid

Φ = phase angle of the sinusoid

(a) 5 + j4

(i) convert to polar form

r = $\sqrt{5^2 + 4^2}$

r = $\sqrt{25 + 16}$

r = $\sqrt{41}$

r = 6.40

Φ = tan⁻¹ ( $\frac{4}{5}$ )

Φ = tan⁻¹ (0.8)

Φ = 38.66°

5 + j4 = 6.40∠38.66°

(ii) Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid

r(t) = 6.40 cos (ωt + 38.66°)

(b) 5 - j4

(i) convert to polar form

r = $\sqrt{5^2 + (-4)^2}$

r = $\sqrt{25 + 16}$

r = $\sqrt{41}$

r = 6.40

Φ = tan⁻¹ ( $\frac{-4}{5}$ )

Φ = tan⁻¹ (-0.8)

Φ = -38.66°

5 - j4 = 6.40∠-38.66°

(ii) Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid

r(t) = 6.40 cos (ωt - 38.66°)

(c) -5 + j4

(i) convert to polar form

r = $\sqrt{(-5)^2 + 4^2}$

r = $\sqrt{25 + 16}$

r = $\sqrt{41}$

r = 6.40

Φ = tan⁻¹ ( $\frac{4}{-5}$ )

Φ = tan⁻¹ (-0.8)

Φ = -38.66°

-5 + j4 = 6.40∠-38.66°

(ii) Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid

r(t) = 6.40 cos (ωt - 38.66°)

(d) -5 - j4

(i) convert to polar form

r = $\sqrt{(-5)^2 + (-4)^2}$

r = $\sqrt{25 + 16}$

r = $\sqrt{41}$

r = 6.40

Φ = tan⁻¹ ( $\frac{-4}{-5}$ )

Φ = tan⁻¹ (0.8)

Φ = 38.66°

-5 - j4 = 6.40∠38.66°

(ii) Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid

r(t) = 6.40 cos (ωt + 38.66°)

3 0

3 years ago

TWO SENTENCES!!!! What are the roles of an engineer?

madam [21]

Answer:

An engineer nomaly cretea things designed by others.They also help repare stuff like cars sorry bout the gammr

Explanation:

5 0

3 years ago

Read 2 more answers

The conditions at the beginning of compression in an Otto engine operating on hot-air standard with k=1.35 and 101.325 kPa, 0.05

Katyanochek1 [597]

Answer:

$P_m=181.42 KPa$

Explanation:

$P_1=101.325 KPa,V_1=0.05m^3,T_1=32C$ ,K=1.35

Clearance is 8%.

Heat added=15 KJ

We know that compression ratio $r=1+\dfrac{1}{C}$

$r=1+\dfrac{1}{0.08}$

r=13.5

$r=\dfrac{V_1}{V_2}$

$13.5=\dfrac{0.05}{V_2}$

$V_2=3.7\times 10^{-3}$

We know that efficiency of otto cycle

$\eta =1-\dfrac{1}{r^{k-1}}$

$\eta =1-\dfrac{1}{13.5^{1.35-1}}$

$\eta =0.56$

$\eta =\dfrac{W}{Q}$

W is the work out put and Q is the heat addition.

$0.56 =\dfrac{W}{15}$

W=8.4 KJ

We know that Work =Mean effective pressure x swept volume.

Here swept volume $V_s=V_1-V_2$

$V_s=0.05-3.7\times 10^{-3}$

$V_s=0.0463 m^3$

Noe by putting the values

Work =Mean effective pressure x swept volume.

$8.4=P_m\times 0.0463$

$P_m=181.42 KPa$

6 0

3 years ago