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Juliette [100K]
3 years ago
5

A ball is attached to a vertical spring. The ball is initially supported at a height y so that the spring is neither stretched n

or compressed. The ball is then released from rest and it falls to a height y - h before moving upward. Consider the following quantities: translational kinetic energy, gravitational potential energy, elastic potential energy. When the ball was at a height y - (h/2), which of the listed quantities has (have) values other than zero joules
Physics
1 answer:
scoundrel [369]3 years ago
3 0

Answer:

All the three quantities will have non zero joules.

Explanation:

At the initial position of rest the system will have only gravitational potential energy while the other 2 quantities will be zero.

when the system reaches a height (y-h) only kinetic energy will be zero while the other 2 quantities will be non zero

At the position of (y-h/2) all the 3 quantities will be non zero.

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A parallel-plate capacitor has square plates that are 7.40 cm on each side and 3.20 mm apart. The space between the plates is co
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Answer:

The energy that can be stored in the capacitor is 239 nJ

Explanation:

We first calculate the capacitance of each material. Let C₁ be the capacitance of pyrex glass and C₂ be the capacitance of polystyrene.

C₁ = κ₁ε₀A/d where κ₁ = dielectric constant of pyrex glass = 5, A = area of plates = L² where L = length of square plate = 7.40 cm = 7.40 × 10⁻² m and d = thickness of pyrex slab = 1.60 mm = 1.60 × 10⁻³ m and ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m

C₁ = κ₁ε₀A/d = κ₁ε₀L²/d = 5 × 8.854 × 10⁻¹² F/m × (7.40 × 10⁻² m)²/1.60 × 10⁻³ m = 2424.2252/1.60 × 10⁻¹¹ F = 1515.14 × 10⁻¹¹ F = 15.2 × 10⁻⁹ F = 15.2 nF

C₂ = κ₂ε₀A/d where κ₂ = dielectric constant of polystyrene = 3, A = area of plates = L² where L = length of square plate = 7.40 cm = 7.40 × 10⁻² m and d = thickness of polystyrene slab = 1.60 mm = 1.60 × 10⁻³ m and ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m

C₁ = κ₁ε₀A/d = κ₁ε₀L²/d = 3 × 8.854 × 10⁻¹² F/m × (7.40 × 10⁻² m)²/1.60 × 10⁻³ m = 1454.5351/1.60 × 10⁻¹¹ F = 909.08 × 10⁻¹¹ F = 9.09 × 10⁻⁹ F = 9.09 nF

Since the capacitors are in series, we find their effective capacitance C from

1/C = 1/C₁ + 1/C₂

C = C₁C₂/(C₁ + C₂)

= 15.2 × 10⁻⁹ F × 9.09 × 10⁻⁹ F/(15.2 × 10⁻⁹ F + 9.09 × 10⁻⁹ F)

= 138.168 × 10⁻¹⁸/24.29 × 10⁻⁹ F

= 5.69 × 10⁻⁹ F

The amount of electrical energy stored in a capacitor is given by W = 1/2CV² where C = capacitance and v = voltage applied. Now C = 5.69 × 10⁻⁹ F and V = 84.0 V for this capacitor

So W = 1/2 × 5.69 × 10⁻⁹ F × 84.0 V

= 238.98 × 10⁻⁹ J

≅ 239 × 10⁻⁹ J

= 239 nJ

So the energy that can be stored in the capacitor is 239 nJ

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Let the cold water go up x degrees.

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Formula

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0.50 * c * x = 1.5 * c * (100 - x)            Divide both sides by c

Solution

0.50 *x = 1.5*(100 - x)                          Remove the brackets.

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Answer

A

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