I believe all of these would be known as specific phobias.
Answer:
El área de la placa es aproximadamente 5102.752 centímetros cuadrados.
Explanation:
Asumamos que el cambio dimensional como consecuencia de la temperatura es pequeña, entonces podemos estimar el área de la placa de cobre en función de la temperatura mediante la siguiente aproximación:
![A_{f} = w\cdot l \cdot [1 + 2\cdot \alpha\cdot (T_{f}-T_{o})]](https://tex.z-dn.net/?f=A_%7Bf%7D%20%3D%20w%5Ccdot%20l%20%5Ccdot%20%5B1%20%2B%202%5Ccdot%20%5Calpha%5Ccdot%20%28T_%7Bf%7D-T_%7Bo%7D%29%5D)
(1)
Donde:
- Ancho de la placa, en centímetros.
- Longitud de la placa, en centímetros.
- Coeficiente de dilatación, en 
.
- Temperatura inicial, en grados Celsius.
- Temperatura final, en grados Celsius.
Si sabemos que 
, 
, 
, 
and 
, entonces el área de la placa a la temperatura final:
![A_{f} = (65\,cm)\cdot (78\,cm)\cdot \left[1+\left(17\times 10^{-6}\,\frac{1}{^{\circ}C} \right)\cdot (400\,^{\circ}C-20\,^{\circ}C)\right]](https://tex.z-dn.net/?f=A_%7Bf%7D%20%3D%20%2865%5C%2Ccm%29%5Ccdot%20%2878%5C%2Ccm%29%5Ccdot%20%5Cleft%5B1%2B%5Cleft%2817%5Ctimes%2010%5E%7B-6%7D%5C%2C%5Cfrac%7B1%7D%7B%5E%7B%5Ccirc%7DC%7D%20%5Cright%29%5Ccdot%20%28400%5C%2C%5E%7B%5Ccirc%7DC-20%5C%2C%5E%7B%5Ccirc%7DC%29%5Cright%5D)
El área de la placa es aproximadamente 5102.752 centímetros cuadrados.
Inertia refers to the tendency of a body to resist changes.
<h3>Inertia</h3>
It is the property of a body to resist change.
A moving body will continue moving until a force acts on it. Unless the acting force is big enough to overcome the inertia of the body, the body will continue moving.
Also, a body at rest will continue resting unless a force big enough to overcome the inertia force is used to displace it.
More on inertia can be found here: brainly.com/question/1358512
The total energy stored in the capacitors is determined as 2.41 x 10⁻⁴ J.
<h3>What is the potential difference of the circuit?</h3>
The potential difference of the circuit is calculated as follows;
U = ¹/₂CV²
where;
- C is capacitance of the capacitor
- V is the potential difference
For a parallel circuit the voltage in the circuit is always the same.
The energy stored in 2.5 μf capacitor is known, hence the potential difference of the circuit is calculated as follows;
U = ¹/₂CV²
2U = CV²
V = √2U/C
V = √(2 x 1.8 x 10⁻⁴ / 2.5 x 10⁻⁶)
V = 12 V
The equivalent capacitance of C1 and C2 is calculated as follows;
1/C = 1/C₁ + 1/C₂
1/C = (1)/(0.9 x 10⁻⁶) + (1)/(16 x 10⁻⁶)
1/C = 1,173,611.11
C = 1/1,173,611.11
C = 8.52 x 10⁻⁷ C
The total capacitance of the circuit is calculated as follows;
Ct = 8.52 x 10⁻⁷ C + 2.5 x 10⁻⁶ C
Ct = 3.35 x 10⁻⁶ C
The total energy of the circuit is calculated as follows;
U = ¹/₂CtV²
U = ¹/₂(3.35 x 10⁻⁶ )(12)²
U = 2.41 x 10⁻⁴ J
Learn more about energy stored in a capacitor here: brainly.com/question/14811408
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