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ArbitrLikvidat [17]
2 years ago
14

The build up of electrons will cause electrical called ?

Engineering
1 answer:
irinina [24]2 years ago
7 0

Answer:

Explanation:

The aufbau principle, from the German Aufbauprinzip (building-up principle), also called the aufbau rule, states that in the ground state of an atom or ion, electrons fill atomic orbitals of the lowest available energy levels before occupying higher levels. For example, the 1s subshell is filled before the 2s subshell is occupied.

feel free to mark me brainliest, if this helps

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valkas [14]
5 is the correct one to choose for this
6 0
2 years ago
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If i eat myself will I get twice as big or disappear completely?
Butoxors [25]
Disappear completely
5 0
2 years ago
A hydraulic jump is induced in an 80 ft wide channel.The water depths on either side of the jump are 1 ft and 10 ft.Please calcu
krek1111 [17]

Answer:

a) 42.08 ft/sec

b) 3366.33 ft³/sec

c) 0.235

d) 18.225 ft

e) 3.80 ft

Explanation:

Given:

b = 80ft

y1 = 1 ft

y2 = 10ft

a) Let's take the formula:

\frac{y2}{y1} = \frac{1}{5} * \sqrt{1 + 8f^2 - 1}

10*2 = \sqrt{1 + 8f^2 - 1

1 + 8f² = (20+1)²

= 8f² = 440

f² = 55

f = 7.416

For velocity of the faster moving flow, we have :

\frac{V_1}{\sqrt{g*y_1}} = 7.416

V_1 = 7.416 *\sqrt{32.2*1}

V1 = 42.08 ft/sec

b) the flow rate will be calculated as

Q = VA

VA = V1 * b *y1

= 42.08 * 80 * 1

= 3366.66 ft³/sec

c) The Froude number of the sub-critical flow.

V2.A2 = 3366.66

Where A2 = 80ft * 10ft

Solving for V2, we have:

V_2 = \frac{3666.66}{80*10}

= 4.208 ft/sec

Froude number, F2 =

\frac{V_2}{g*y_2} = \frac{4.208}{32.2*10}

F2 = 0.235

d) El = \frac{(y_2 - y_1)^3}{4*y_1*y_2}

El = \frac{(10-1)^3}{4*1*10}

= \frac{9^3}{40}

= 18.225ft

e) for critical depth, we use :

y_c = [\frac{(\frac{3366.66}{80})^2}{32.2}]^1^/^3

= 3.80 ft

7 0
3 years ago
Read 2 more answers
A converging-diverging nozzle has an area ratio of 5.9. (1) Determine the (P0/Pt) values corresponding to the 1st, 2nd, and 3rd
nata0808 [166]

Answer:

Check the explanation

Explanation:

The Total pressure is the overall of fixed or static pressure p, the dynamic pressure q, as well as gravitational head. Total pressure can also be referred to as the measure of the overall energy of the airstream, and is the same to static pressure plus velocity pressure.

kindly check the step by step solution in the attached image below to Determine the (P0/Pt) values corresponding to the 1st, 2nd, and 3rd critical points.

5 0
3 years ago
Design a 7.5-V zener regulator circuit using a 7.5-V zener specified at 10mA. The zener has an incremental resistance of rZ = 30
hram777 [196]

Answer:

The answer is given in the explanation.

Explanation:

The circuit is as indicated in the attached figure.

From the analytical description the zener voltage is given as

V_z=V_z_o+I_zr_z

Here

Vzo is the voltage at which the slope of 1/rz intersects the voltage axis it is equal to knee voltage.

The equivalent model is shown in the attached figure.

From the above equation, Vzo is calculated as

V_z_o=V_z-I_zr_z

Here Vz is given as 7.5 V

Iz is given as 10 mA

rz is given as 30 Ω

Thus the Vzo is given as

V_z_o=V_z-I_zr_z\\V_z_o=7.4-30*10*10^{-3}\\V_z_o=7.5-0.3\\V_z_o=7.2 V

The value of I_L is given as 5 mA

Now the expression of current is as

I=I_z+I_L\\I=10mA+5mA\\I=15 mA

Now the resistance is calculated as

R=\dfrac{V-Vo}{I}\\R=\dfrac{10-7.2}{15*10^{-3}}\\R=186.66

So the value of resistance is 186.66 Ω.

Considering the supply voltage is increased by 10%

V is 10-10%*10=10+1=11 so the

R=\dfrac{V-Vo}{I}\\186.66=\dfrac{11-V_o}{15*10^{-3}}\\V_o=8.2 V

Considering the supply voltage is decreased by 10%

V is 10-10%*10=10-1=9 so the

R=\dfrac{V-Vo}{I}\\186.66=\dfrac{9-V_o}{15*10^{-3}}\\V_o=6.2 V

Now if the supply voltage is 10% high and the value of Load is removed i.e I=Iz only which is 10mA

so

R=\dfrac{V-Vo}{I'}\\186.66=\dfrac{11-V_o}{10*10^{-3}}\\V_o=9.13 V

Now the largest load current thus that the supply voltage is 10% low and the current of zener is knee current thus

V_z_o=V_z-I_zr_z\\V_z_o=7.5-30*0.5*10^{-3}\\V_z_o=7.5-0.015\\V_z_o=7.485 V

R=\dfrac{V-Vo}{I'}\\186.66=\dfrac{9-7.485}{I}\\I=10.71 mA

The load voltage is 7.485 V

8 0
2 years ago
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