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The change in potential along a path from C to D due to a small charged sphere is 900 V.
Given:
Charge, Q = 3 nC = 3 × 10⁻⁹ C
Distance between the sphere and point C, r₁ = 0.02 m
Distance between the sphere and point D, r₂ = 0.06 m
Calculation:
We know that the electric potential is given as:
V = k Q/r - (1)
where, V is the electric potential
k is Coulomb's force constant
Qis the charge on the sphere
r is the separation distance
The electric potential at point C due to charged sphere can be given as:
V₁ = k Q/r₁
= (9×10⁹ Nm²/C²) [(3 × 10⁻⁹ C)/(0.02 m)]
= 1350 V
The electric potential at point D due to charged sphere can be given as:
V₂ = k Q/r₂
= (9×10⁹ Nm²/C²) [(3 × 10⁻⁹ C)/(0.06 m)]
= 450 V
Now, the change in potential along the path from C to D can be calculated as:
ΔV = V₂ - V₁
= 450 V - 1350 V
= -900 V
The negative sign indicates that the work is done against the electric field in moving the charge from C to D.
Therefore, the change in potential along a path from C to D is 900 V against the direction of the electric field.
Learn more about the electric potential here:
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Answer:
El gasto de gasto es de aproximadamente 0.0273 pies cúbicos por segundo.
Explanation:
El gasto es el flujo volumétrico de gasolina (), medido en pies cúbicos por segundo, que sale de la manguera. Asumiendo que la velocidad de salida es constante, tenemos que el gasto a través de la manguera es:
(1)
Donde:
- Diámetro de la manguera, medido en pies.
- Velocidad medida de salida, medida en pies por segundo.
Si sabemos que y
, entonces el gasto de gasolina es:
El gasto de gasto es de aproximadamente 0.0273 pies cúbicos por segundo.