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3 years ago

List four conditions that may result in the emission of electrons from a conductor

Physics

2 answers:

QveST [7]3 years ago

8 0

There are 4 ways in which electrons are emitted from the conductor.

i. Thermionic emission

ii. Electric field electron emission

iii. Photoelectric emission

iv. Secondary emission

In thermionic emission large amount of external energy in the form of heat is supplied to release free electrons from the metal.

In electric field electron emission, electrons are emitted from the metal surface when the metals are placed in a very strong electric field.

During photoelectric emission, light is absorbed by the metals and this provides energy to the valence electrons which break their bond with the parent atom and which are then released from the atom.

Valence electrons do have some kinetic energy, but they don't have enough energy to escape from the atom. During secondary emission, a high-speed electron is bombarded with an atom, which provides the energy for the valence electrons to break their bonds with their parent atom which are then released from the atom.

NNADVOKAT [17]3 years ago

3 0

Answer:

Conditions that result in the emission of electrons from a conductor:

Heating the conductor to a suitable temperature

Exposing the conductor to a strong light

Subjecting the conductor to a very high applied voltage

Subjecting the conductor to high-speed electrons from another source

Explanation:

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Help!!!, combination circuits, Physics

Kaylis [27]

Current and voltage on each resistor:

$I_1 = 3.98 A, V_1 = 3.98 V$

$I_2=0.015 A, V_2 = 0.075 V$

$I_3 = 0.4 A, V_3 = 0.4 V$

$I_4 = 0.385 A, V_4 = 0.77 V$

$I_5 = 0.585 A, V_5 = 1.17 V$

$I_6 = 3.01 A, V_6 = 6.02 V$

$I_7 = 0.97 A, V_7 = 4.85 V$

Explanation:

In order to solve the circuit, we first have to find the equivalent resistance of the whole circuit, then the total current, and then we can proceed finding the current and the voltage for each resistor.

We start by calculating the equivalent resistance of resistors 2 and 3, which are in parallel:

$R_{23}=\frac{R_2R_3}{R_2+R_3}=\frac{(5)(1)}{5+1}=0.833\Omega$

This resistor is in series with resistor 4, so:

$R_{234}=R_{23}+R_4=0.833+2.0=2.833\Omega$

This resistor is in parallel with resistor 5, therefore:

$R_{2345}=\frac{R_{234}R_5}{R_{234}+R_5}=\frac{(2.833)(2.0)}{2.833+2.0}=1.172\Omega$

This resistor is in series with resistor 7, so:

$R_{23457}=R_{2345}+R_7=1.172+5.0=6.172\Omega$

This resistor is in parallel with resistor 6, so:

$R_{234567}=\frac{R_{23457}R_6}{R_{23457}+R_6}=\frac{(6.172)(2.0)}{6.172+2.0}=1.510\Omega$

Finally, this combination is in series with resistor 1:

$R_{eq}=R_1+R_{234567}=1.0+1.510=2.510\Omega$

We finally found the equivalent resistance of the circuit. Now we can find the total current in the circuit, which is also the current flowing through resistor 1:

$I_1=\frac{V}{R_{eq}}=\frac{10}{2.510}=3.98 A$

And we can also find the potential difference across resistor 1:

$V_1=I_1 R_1=(3.98)(1.0)=3.98 V$

This means that the voltage across resistor 6 is

$V_6=V-V_1=10-3.98=6.02 V$

And so, the current on resistor 6 is

$I_6=\frac{V_6}{R_6}=\frac{6.02}{2.0}=3.01 A$

The current flowing in the whole part of the circuit containing resistors 2,3,4,5,7, and therefore through resistor 7, is

$I_7=I-I_6=3.98-3.01=0.97 A$

And so the voltage across resistor 7 is

$V_7=I_7 R_7=(0.97)(5.0)=4.85 V$

The voltage across resistor 5 is

$V_5 = V_6 - V_7 = 6.02 - 4.85 =1.17 V$

And so the current is

$I_5 = \frac{V_5}{R_5}=\frac{1.17}{2.0}=0.585 A$

The current through resistor 4 is

$I_4 = I_7 - I_5 = 0.97-0.585 = 0.385 A$

And therefore its voltage is

$V_4=I_4 R_4 = (0.385)(2.0)=0.77 V$

So, the voltage through resistor 3 is

$V_3=V_5-V_4=1.17-0.77=0.4 V$

And the current is

$I_3=\frac{V_3}{R_3}=\frac{0.4}{1.0}=0.4 A$

Finally, the current through resistor 2 is

$I_2=I_4-I_3=0.5-0.385=0.015 A$

And so its voltage is

$V_2=I_2R_2=(0.015)(5.0)=0.075 V$

Learn more about current and voltage:

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4 0

4 years ago

Need help on this please

Likurg_2 [28]

Build up of pressure between tectonic plates .

8 0

3 years ago

A baby carriage is rolling down a hill. Take into consideration, the carriage has 90J of kinetic energy and the mass of the carr

Sphinxa [80]

Answer:

YES. 6√10

Explanation:

apply E=1/2mv^2 and find the velocity of carriage which is 6√10 ms-1

as our velocity is higher than that of cart. we can catch the carriage

7 0

3 years ago

The formula $F = \frac{9}{5} C + 32$ can be used to convert temperatures between degrees Fahrenheit ($F$) and degrees Celsius ($

slamgirl [31]

Answer:

-30° C

Explanation:

Data provided in the problem:

The formula for conversion as:

F = (9/5)C + 32

Now,

for the values of F = -22 , C = ?

Substituting the value of F in the above formula, we get

-22 = (9/5)C + 32

-22 - 32 = (9/5)C

(9/5)C = - 54

C = - 54 × (5/9)

C = - 30 °

Hence, -22 Fahrenheit equals to -30°C

7 0

3 years ago

The area of the piston to the master cylinder in a hydraulic braking system of a car is 0.6 square inches. If a force of 5.6 lb

balu736 [363]

Answer:

16.8 lb is the force on the brake pad of one wheel.

Explanation:

Force applied on the piston = $F_1=5.6 lb$

Area of the piston = $A_1=0.6 inches^2$

Force applied on the brakes = $F_2$

Area of the brakes = $A_2=1.8 inches^2$

Applying Pascal's law: 'For an incompressible fluid pressure at one surface is equal to the pressure at other surface'.

$\frac{F_1}{A_2}=\frac{F_2}{A_2}$

$F_2=\frac{5.6 lb\times 1.8 inhes^2}{0.6 inches^2}=16.8 lb$

16.8 lb is the force on the brake pad of one wheel.

5 0

4 years ago