The work done to transport an electron from the positive to the negative terminal is 1.92×10⁻¹⁹ J.
Given:
Potential difference, V = 1.2 V
Charge on an electron, e = 1.6 × 10⁻¹⁹ C
Calculation:
We know that the work done to transport an electron from the positive to the negative terminal is given as:
W.D = (Charge on electron)×(Potential difference)
= e × V
= (1.6 × 10⁻¹⁹ C)×(1.2 V)
= 1.92 × 10⁻¹⁹ J
Therefore, the work done in bringing the charge from the positive terminal to the negative terminal is 1.92 × 10⁻¹⁹ J.
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Responder:
<h2>
490 julios
</h2>
Explicación:
Se dice que el trabajo se realiza cuando una fuerza aplicada a un objeto hace que el objeto se mueva a través de una distancia. El trabajo realizado por un cuerpo se expresa mediante la fórmula;
Workdone = Fuerza * Distancia
Como Fuerza = masa * aceleración,
Workdone = masa * aceleración * distancia
Masa dada = 5.0kg, aceleración = 2.0m / s² d =?
Para obtener d, usaremos una de las leyes del movimiento,
d = ut + 1 / 2at²
u = 0 (ya que el cuerpo acelera desde el reposo) yt = 7.0s
d = 0 + 1/2 (2) (7) ²
d = 49m
Workdone = 5 * 2 * 49
Workdone = 490 Julios
Answer: B. store electric charges.
Explanation: I JUST TOOK THE PF EXAM AND I GOT IT CORRECT!!!!
Answer:
Explanation:
Start with Carbon and assume we only get 1 sugar molecule from the process.
you have 6 carbons in the sugar on the right, so you need 6 carbons on the left which only come from CO₂
6 CO₂
you have 12 hydrogen atoms in the sugar on the right, so you need 12 hydrogen atoms on the left which only come from H₂O. At 2 hydrogen atoms per water molecule means you need 6 waters.
6 CO₂ + 6 H₂O → 1 C₆H₁₂O₆
you are supplied with 12 oxygen from the CO₂ and 6 oxygen from the H₂O, but you only need 6 oxygen for the sugar. That means there are 12 oxygen remaining which will become 6 O₂ molecules
6 CO₂ + 6 H₂O → 1 C₆H₁₂O₆ + 6 O₂
