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3 years ago

A spring with k = 500 N/m stores 704 J. How far is it extended from the equilibrium position

Physics

1 answer:

kaheart [24]3 years ago

4 0

Given that:

k = 500 n/m,

work (W) = 704 J

spring extension (x) = ?

we know that,

Work = (1/2) k x²

704 = (1/2) × 500 × x²

x = 1.67 m

A spring stretched for 1.67 m distance.

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A pendulum of 50 cm long consists of small ball of 2kg starts swinging down from height of 45cm at rest. the ball swings down an

Ket [755]

Assuming that all energy of the small ball is transferred to the bigger ball upon impact, then we can say that:

Potential Energy of the small ball = Kinetic Energy of the bigger ball

Potential Energy = mass * gravity * height

Since the small ball start at 45 cm, then the height covered during the swinging movement is only:

height = 50 cm – 45 cm = 5 cm = 0.05 m

Calculating for Potential Energy, PE:

PE = 2 kg * 9.8 m / s^2 * 0.05 m = 0.98 J

Therefore, maximum kinetic energy of the bigger ball is:

Max KE = PE = 0.98 J

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3 years ago

The Lamborghini Huracan has an initial acceleration of 0.80g. Its mass, with a driver, is 1510 kg. If an 80 kg passenger rode al

irina [24]

Answer:

a = 15.1 g

Explanation:

The relation between mass and acceleration is given by :

$m\propto \dfrac{1}{a}$

If a₁ = 0.80g, m₁ = 1510 kg, m₂ = 80 kg, we need to find a₂

So,

$\dfrac{m_1}{m_2}=\dfrac{a_2}{a_1}\\\\a_2=\dfrac{a_1m_1}{m_2}\\\\a_2=\dfrac{0.8g\times 1510}{80}\\\\a_2=15.1g$

So, the car's acceleration would be 15.1 g.

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3 years ago

A 5.00-V battery charges the parallel plates in a capacitor, with a plate area of 865 mm2 and an air-filled separation of 3.00 m

Westkost [7]

Answer:

W = 3.21x10⁻¹¹ J

Explanation:

The work required to separate the plates can be calculated using the following equation:

$W = U_{2} - U_{1} = \frac{1}{2}(C_{2}V_{2}^{2} - C_{1}V_{1}^{2})$

Where:

U₂: is the final stored energy

U₁: is the initial stored energy

C₂: is the final capacitance

C₁: is the initial capacitance

V₁: is the initial potential difference = 5.00 V

V₂: is the final potential difference

The initial and final capacitance is:

$C_{1} = \epsilon_{0}*\frac{A}{d_{1}}$

Where:

ε₀: is the vacuum permittivity = 8.85x10⁻¹² C²/(N*m²)

d: is the initial distance = 3.00 mm = 3.00x10⁻³ m

A: is the plate area = 865 mm² = 8.65x10⁻⁴ m²

$C_{1} = \epsilon_{0}*\frac{A}{d_{1}} = 8.85 \cdot 10^{-12} C^{2}/(N*m^{2})*\frac{8.65 \cdot 10^{-4} m^{2}}{3.00 \cdot 10^{-3} m} = 2.55 \cdot 10^{-12} F$

Similarly, C₂ is:

$C_{2} = \epsilon_{0}*\frac{A}{d_{2}} = 8.85 \cdot 10^{-12} C^{2}/(N*m^{2})*\frac{8.65 \cdot 10^{-4} m^{2}}{3.00 + 3.00 \cdot 10^{-3} m} = 1.28 \cdot 10^{-12} F$

Now, V₂ can be calculated by finding the initial charge (q₁):

$q_{1} = C_{1}V_{1} = 2.55 \cdot 10^{-12} F*5.00 V = 1.28 \cdot 10^{-11} C$

Since, q₁ is equal to q₂, V₂ is:

$V_{2} = \frac{q_{2}}{C_{2}} = \frac{1.28 \cdot 10^{-11} C}{1.28 \cdot 10^{-12} F} = 10 V$

Finally, we can find the work:

$W = \frac{1}{2}(C_{2}V_{2}^{2} - C_{1}V_{1}^{2}) = \frac{1}{2}(1.28 \cdot 10^{-12} F*(10 V)^{2} - 2.55 \cdot 10^{-12} F(5.00 V)^{2}) = 3.21 \cdot 10^{-11} J$