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3 years ago

If i build thing a and thing a builds thing b did i build thing b

Engineering

2 answers:

lianna [129]3 years ago

8 0

Answer:

No, I don't think so.

Explanation:

It depends on if you programmed thing a to build thing b.

postnew [5]3 years ago

3 0

Answer:

Explanation:

If you build thing "a" and thing "a" builds thing "b" you <u>indirectly</u> build thing "b".

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Write multiple if statements:

lora16 [44]

Zrizorzlzfxxxgoxxxxpgxtoxxxhxuxyf

3 0

2 years ago

A 20.0 µF capacitor is charged to a potential difference of 800 V. The terminals of the charged capacitor are then connected to

Sergeu [11.5K]

Answer:

a) Q_initial = 16 * 10^-3 C

b) V_1 = V_2 = (16/3) * 10^2 V

c) E = 64/15 J

d) dE = 32/15 J of decrease

Explanation:

Given:

- Capacitor 1, C_1 = 20.0 uF

- Capacitor 2, C_2 = 10.0 uF

- Charged with P.d V = 800 V

Find:

a) the original charge of the system,

(b) the ﬁnal potential difference across each capacitor

(d) the decrease in energy when the capacitors are connected.

Solution:

- The initial charge in the circuit is the one carried by the first charged capacitor.

Q_initial = C_1*V

Q_initial = 20*10^-6 * 800

Q_initial = 16 * 10^-3 C

- After charging the other capacitor, we know that the total charge is conserved among two capacitor:

Q_initial = Q_1 + Q_2

- We also know that potential difference across two capacitor is also same.

V_1 = V_2 = Q_1 / C_1 = Q_2 / C_2

- Using the two equations and solve for charge Q_2:

Q_2 = Q_1*C_2/C_1

Q_2 = Q_1*10/20 = 0.5*Q_1

- using conservation of charge:

Q_initial = 1.5*Q_1

Q_1 = 16*10^-3 / 1.5 = 10.67*10^-3 C

- Hence the Voltage across each capacitor is:

V_2 = V_1 = Q_1 / C_1

= 10.67*10^-3 / 20*10^-6

= (16/3) * 10^2 V

- The energy in the system is:

E = 0.5*C_eq*V^2

Where, C_eq is the equivalent capacitance of paralle circuit.

E = 0.5*(20+10)*10^-6 *((16/3) * 10^2)^2

E = 64/15 J

- The decrease in energy of the capacitors is:

dE = E_initial - E_final

Where, E_initial is due to charging of the C_1 only:

dE = 0.5*10^-6*20*800^2 - (64/15)

dE = 32/5 - 64/15 = 32/15 J

5 0

3 years ago

.a. What size vessel holds 2 kg water at 80°C such that 70% is vapor? What are the pressure and internal energy? b. A 1.6 m3 ves

vesna_86 [32]

Answer:

Part a: The volume of vessel is 4.7680 $m^3$ and total internal energy is 3680 kJ.

Part b: The quality of the mixture is 90.3% or 0.903, temperature is 120 °C and total internal energy is 4660 kJ.

Explanation:

Part a:

As per given data

m=2 kg

T=80 °C =80+273=353 K

Dryness=70% vapour =0.7

<em>From the steam tables at 80 °C</em>

Specific volume of saturated vapours=v_g=3.40527 $m^3/kg$

Specific volume of saturated liquid=v_f=0.00102 $m^3/kg$

Now the relation of total specific volume for a specific dryness value is given as

$v=v_f+x(v_g-v_f)$

Substituting the values give

$v=v_f+x(v_g-v_f)\\v=0.00102+0.7(3.40527-0.00102)\\v_f=2.38399 m^3/kg$

Now the volume of vessel is given as

$v=\frac{V}{m}\\V=v \times m\\V=2.38399 \times 2\\V=4.7680 m^3$

So the volume of vessel is 4.7680 $m^3$ .

Similarly for T=80 and dryness ratio of 0.7 from the table of steam

Pressure=P=47.4 kPa

Specific internal energy is given as u=1840 kJ/kg

So the total internal energy is given as

$u=\frac{U}{m}\\U=u \times m\\U=1840 \times 2\\U=3680 kJ$

The total internal energy is 3680 kJ.

So the volume of vessel is 4.7680 $m^3$ and total internal energy is 3680 kJ.

Part b

Volume of vessel is given as 1.6

mass is given as 2 kg

Pressure is given as 0.2 MPa or 200 kPa

Now the specific volume is given as

$v=\frac{V}{m}\\v=\frac{1.6}{2}\\v=0.8 m^3/kg$

So from steam tables for Pressure=200 kPa and specific volume as 0.8 gives

Temperature=T=120 °C

Quality=x=0.903 ≈ 90.3%

Specific internal energy =u=2330 kJ/kg

The total internal energy is given as

$u=\frac{U}{m}\\U=u \times m\\U=2330 \times 2\\U=4660 kJ$

So the quality of the mixture is 90.3% or 0.903, temperature is 120 °C and total internal energy is 4660 kJ.

5 0

3 years ago

Can someone answer plz!! It’s 24 points

fgiga [73]

Explanation:

750 microvolt is your answer

please mark as brilliant

3 0

2 years ago

Read 2 more answers

In C++ the declaration of floating point variables starts with the type name float or double, followed by the name of the variab

zubka84 [21]

Answer:

The given grammar is :

S = T V ;

V = C X

X = , V | ε

T = float | double

C = z | w

Nullable variables are the variables which generate ε ( epsilon ) after one or more steps.

From the given grammar,

Nullable variable is X as it generates ε ( epsilon ) in the production rule : X -> ε.

No other variables generate variable X or ε.

So, only variable X is nullable.

First of nullable variable X is First (X ) = , and ε (epsilon).

L.H.S.

The first of other varibles are :

First (S) = {float, double }

First (T) = {float, double }

First (V) = {z, w}

First (C) = {z, w}

R.H.S.

First (T V ; ) = {float, double }

First ( C X ) = {z, w}

First (, V) = ,

First ( ε ) = ε

First (float) = float

First (double) = double

First (z) = z

First (w) = w

Follow of nullable variable X is Follow (V).

Follow (S) = $

Follow (T) = {z, w}

Follow (V) = ;

Follow (X) = Follow (V) = ;

Follow (C) = , and ;

Explanation:

4 0

3 years ago