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erma4kov [3.2K]
3 years ago
13

A simple pendulum with a length of 2 m oscillates on the Earth's surface. What is the period of oscillations?

Physics
1 answer:
katrin [286]3 years ago
5 0

Answer:

Period, T = 27.84 seconds

Explanation:

Given the following data;

Length of pendulum = 2m

We know that acceleration due to gravity on earth is equal to 9.8 m/s²

To find the period of oscillation of the simple pendulum;

Mathematically, it is given by the formula;

Period, T = 2 \pi \sqrt {lg}

Substituting into the equation, we have;

Period, T = 2 * 3.142 * \sqrt {2*9.8}

Period, T = 6.284 * \sqrt {19.6}

Period, T = 6.284 * 4.43

Period, T = 27.84 seconds

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An asteroid is on a collision course with Earth. An astronaut lands on the rock to bury explosive charges that will blow the ast
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Answer:

The maximum radius the asteroid can have for her to be able to leave it entirely simply by jumping straight up is approximately 1782.45 meters

Explanation:

Whereby the height the astronaut can jump on Earth = 0.500 m, we have the following kinematic equation;

v² = u² - 2·g·h

Where;

v = The final velocity

u = The initial velocity

g = The acceleration due to gravity ≈ 9.8 m/s²

h = The height she jumps

At the maximum height, h_{max} = 0.500 m, she jumps, v = 0, therefore, we have;

0² = u² - 2·g·h_{max}

u² = 2 × 9.8 × 0.5 = 9.8

u = √9.8 ≈ 3.13

u = 3.13 m/s

Her initial jumping velocity ≈ 3.13 m/s

Escape velocity, v_e = \sqrt{\dfrac{2 \cdot G \cdot M}{r} }

Where;

M = The mass of the asteroid

G = The Universal gravitational constant = 6.67408 × 10⁻¹¹ m³/(kg·s²)

r = The radius of the asteroid

The average density of the Earth = 5515 kg/m³

The mass of the asteroid, M = Density × Volume = 5515 kg/m³× 4/3 × π × r³

The escape velocity, she has, v_e ≈ 3.13 m/s is therefore;

3.13 = \sqrt{\dfrac{2 \times 6.67408 \times 10^{-11} \times 5515 \times \frac{4}{3} \times \pi \times r^3}{r} } = r \times \sqrt{3.084 \times 10^{-6}}

r = \dfrac{3.13}{ \sqrt{3.084 \times 10^{-6}}} \approx 1782.45

Therefore, the maximum radius of the asteroid can have for her jumping velocity to be equal to the escape velocity for her to be able to leave it entirely simply by jumping straight up = r ≈ 1782.45 meters.

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