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3 years ago

Is energy absorbed or given off when an electron and a hole of a semi-conductor crystal

Physics

1 answer:

Nina [5.8K]3 years ago

7 0

Answer:

i have no clue XD

Explanation:

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PROBLEMA DI FISICA:il volume di un attizzatoio da camino in ferro, a temperatura ambiente (20°C), è 12,2 cm cubi. A contatto con

lesantik [10]

Responder:

20.3 ° C

Explicación:

Según la ley de Charles: cuando la presión sobre una muestra de gas seco se mantiene constante, la temperatura y el volumen estarán en proporción directa.

Paso uno:

datos dados

Temperatura T1 = 20 ° C

Temperatura T2 =?

Volumen V1 = 12.2 cm ^ 3

Volumen V2 = 12.4 cm ^ 3

Aplicar la relación temperatura y volumen

$\frac {V_{1}}{T_{1}}=\frac {V_{2}}{T_{2}}$

sustituyendo tenemos

$\frac {12.2}{20}=\frac {12.4}{T_{2}}\\\\$

Cruz multiplicar tenemos

$T_2=\frac{20*12.4}{12.2} \\\\T_2=\frac{248}{12.2}\\\\T_2=20.3$

Temperatura delle braci 20.3°C

4 0

3 years ago

find the period of a simple pendulum of 1m length placed on earth and on moon g on moon =1.67m/s² g on earth=10m/s²

Ierofanga [76]

Answer:

$T_{m }$ = 4.86 s

$T_{e}$ = 1.98 s

Explanation:

Given:

Length = l = 1 m

Acceleration due to gravity of moon = $g_{m}$ = 1.67 m/s²

Acceleration due to gravity of Earth = $g_{e}$ = 10 m/s²

Required:

Time period = T = ?

Formula:

T = 2π $\sqrt{\frac{l}{g} }$

Solution:

For moon

Putting the givens,

T = 2(3.14) $\sqrt{\frac{1}{1.67} }$

T = 6.3 $\sqrt{0.6}$

T = 6.3 × 0.77

T = 4.86 sec

For Earth,

Putting the givens

T = 2π $\sqrt{\frac{1}{10} }$

T = 2(3.14) $\sqrt{0.1}$

T = 6.3 × 0.32

T = 1.98 sec

3 0

3 years ago

What process is used to release energy in nuclear power plants

Anika [276]

.........Nucler fission.... .

5 0

3 years ago

Read 2 more answers

Can anyone explain how tides work

Mila [183]

The position of the sun and the moon affect how high the tide is

6 0

3 years ago

Read 2 more answers

A mountain climber, in the process of crossing between two cliffs by a rope, pauses to rest. She weighs 514 N. As the drawing sh

n200080 [17]

Answer:

Part a)

$T_L = 155.4 N$

Part b)

$T_R = 379 N$

Explanation:

As we know that mountain climber is at rest so net force on it must be zero

So we will have force balance in X direction

$T_L cos65 = T_R cos80$

$T_L = 0.41 T_R$

now we will have force balance in Y direction

$mg = T_L sin65 + T_Rsin80$

$514 = 0.906T_L + 0.985T_R$

Part a)

so from above equations we have

$514 = 0.906T_L + 0.985(\frac{T_L}{0.41})$

$514 = 3.3 T_L$

$T_L = 155.4 N$

Part b)

Now for tension in right string we will have

$T_R = \frac{T_L}{0.41}$

$T_R = 379 N$

3 0

3 years ago