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lara [203]
3 years ago
14

Suppose that an object travels from one point in space to another. Make a comparison between the magnitude of the displacement a

nd the distance traveled by this object.
Physics
1 answer:
Rashid [163]3 years ago
3 0

Answer:

- Distance is a scalar quantity, defined as the total amount of space covered by an object while moving between the final position and the initial position. Therefore, it depends on the path the object has taken: the distance will be minimum if the object has travelled in a straight line, while it will be larger if the object has taken a non-straight path.

- Displacement is a vector quantity, whose magnitude is equal to the distance (measured in a straight line) between the final position and the initial position of the object. Therefore, the displacement does NOT depend on the path taken, but only on the initial and final point of the motion.

If the object has travelled in a straight path, then the displacement is equal to the distance. In all other cases, the distance is always larger than the displacement.

A particular case is when an object travel in a circular motion. Assuming the object completes one full circle, we have:

- The distance is the circumference of the circle

- The displacement is zero, because the final point corresponds to the initial point

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A parallel-plate, air-gap capacitor has a capacitance of 0.14 mu F. The plates are 0.5 mm apart, What is the area of each plate?
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Answer:

7.9060 m²

8.57 Volts

5.142×10⁻⁶ Joule

1.2×10⁻⁶ Coulomb

Explanation:

C = Capacitance between plates = 0.14 μF = 0.14×10⁻⁶ F

d = Distance between plates = 0.5 mm = 0.5×10⁻³ m

Q = Charge = 1.2 μC = 1.2×10⁻⁶ C

ε₀ = Permittivity = 8.854×10⁻¹² F/m

Capacitance

C=\frac{\epsilon_{0}A}{d}\\\Rightarrow A=\frac{Cd}{\epsilon_{0}}\\\Rightarrow A=\frac{0.14\times 10^{-6}\times 0.5\times 10^{-3}}{8.854\times 10^{-12}}\\\Rightarrow A=7.9060\ m^2

∴ Area of each plate is 7.9060 m²

Voltage

V=\frac{Q}{C}\\\Rightarrow V=\frac{1.2\times 10^{-6}}{0.14\times 10^{-6}}\\\Rightarrow V=8.57\ Volts

∴ Potential difference between the plates if the capacitor is charged to 1.2 μC  is 8.57 Volts.

Energy stored

E=0.5CV²

⇒E = 0.5×0.14×10⁻⁶×8.57²

⇒E = 5.142×10⁻⁶ Joule

∴ Stored energy is 5.142×10⁻⁶ Joule

Charge

Q = CV

⇒Q = 0.14×10⁻⁶×8.57

⇒Q = 1.2×10⁻⁶ C

∴ Charge the capacitor carries before a spark occurs between the two plates is 1.2×10⁻⁶ Coulomb

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Answer:

0.775 m

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