In a science museum, a 130 kg brass pendulum bob swings at the end of a 14.4 m -long wire. The pendulum is started at exactly 8:00 a.m. every morning by pulling it 1.7 m to the side and releasing it. Because of its compact shape and smooth surface, the pendulum's damping constant is only 0.010kg/s. You may want to review (Pages 405 - 407) . Part A At exactly 12:00 noon, how many oscillations will the pendulum have completed

Answers

Answer 1
Answer:

Answer:

The time in which the pendulum does a complete revolution is called the period of the pendulum.

Remember that the period of a pendulum is written as:

T = 2*pi*√(L/g)

where:

L = length of the pendulum

pi = 3.14

g = 9.8 m/s^2

Here we know that  L = 14.4m

Then the period of the pendulum will be:

T = 2*3.14*√(14.4m/9.8m/s^2) = 7.61s

So one complete oscillation takes 7.61 seconds.

We know that the pendulum starts moving at 8:00 am

We want to know 12:00 noon, which is four hours after the pendulum starts moving.

So, we want to know how many complete oscillations happen in a timelapse of 4 hours.

Each oscillation takes 7.61 seconds.

The total number of oscillations will be the quotient between the total time (4 hours) and the period.

First we need to write both of these in the same units, we know that 1 hour = 3600 seconds

then:

4 hours = 4*(3600 seconds) = 14,400 s

The total number of oscillations in that time frame is:

N = 14,400s/7.61s = 1,892.25

Rounding to the next whole number, we have:

N = 1,892

The pendulum does 1,892 oscillations between 8:00 am and 12:00 noon.

Answer 2
Answer:

Final answer:

The question involves the concept of a simple pendulum whose number of swings is largely influenced by its length and the acceleration due to gravity. By determining the period of the pendulum, one can figure out the number of oscillations over a given time period. The pendulum's damping constant is negligible in determining the number of oscillations.

Explanation:

The subject of this question involves understanding the concept of a simple pendulum and how it relates to harmonic motion. It is widely known that the mass of the pendulum does not influence the oscillations but rather the length of the pendulum wire and acceleration due to gravity are paramount.

First, the necessary step toward calculating the number of swings would be to calculate the period of the pendulum's oscillation. This is given by the formula T=2*π*sqrt(L/g), where L is the length of the pendulum (14.4m) and g is the acceleration due to gravity (~9.81m/s²). Substituting these values will give us the period, T, in seconds.

The pendulum starts swinging at 8:00 am and at 12:00 noon, 4 hours or 14400 seconds will have passed. Therefore the number of oscillations will be calculated by dividing the total time by one period of oscillation.

It is crucial to note that the damping in this instance is quite small and would not significantly affect the number of oscillations.

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An small asteroid with a mass of 6.25 x 1014 kg (almost 10 billion times less mass than the Earth) creates a force of 360 N on a nearby spacecraft that has a mass of 7.20 x 105 kg, how far apart are these two objects in km?

Answers

The Objects separated by gravitational force of 360 N, masses 6.25 × 10¹⁴ kg and 7.20 × 10⁵ kg. Applying Newton's law with G = 6.67430 × 10⁻¹¹ m³ kg⁻¹ s⁻² yields distance is 2.79139 km.

We can apply Newton's law of universal gravitation to calculate the distance between the two objects. The formula takes the form:

F = (G ⋅ m₁ ⋅ m₂) / r²

Where:

F represents the force between the two objects

G stands for the gravitational constant (6.67430 × 10⁻¹¹ m³ kg⁻¹ s⁻²)

m₁ and m₂ denote the masses of the two objects

r indicates the distance between the centers of the two objects

Given that the force F = 360 N, m₁ = 6.25 × 10¹⁴ kg, and m₂ = 7.20 × 10⁵ kg, we can solve for r:

r² = (G ⋅ m₁ ⋅ m₂) / F

r = √((G ⋅ m₁ ⋅ m₂) / F)

Now, substituting the values and solving for r:

r = √(((6.67430 × 10⁻¹¹ m³ kg⁻¹ s⁻²) ⋅ (6.25 × 10¹⁴ kg) ⋅ (7.20 × 10⁵ kg)) / 360 N)

r ≈ 2791.39 m

Finally, converting the distance from meters to kilometers:

r ≈ 2.79139 km

Consequently, the two objects are approximately 2.79139 kilometers apart.

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The distance between the asteroid and the spacecraft is 2.3 km.

Timers at a swim meet used four different clocks to time an event. Which recorded time is the least precise?a. 48.9107 s
b. 48.91 s
c. 48.910 s
d. 48.9 s

Answers

Answer:

Least precise recorded time is 48.9 seconds.

Explanation:

Here, we need to write the recorded time that is least precise. The given options are (a) 48.9107 s  (b) 48.91 s  (c) 48.910 s  (d) 48.9 s. The precision in any measurement is defined as closeness in any measurement.

First option is most precise as it is precised to four decimal places. Second option is precised to two decimal places. Third option is precised to 3 decimal places. But in option fourth, the recorded time is precised to one decimal place.

Hence, the correct option is (d).

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Explanation:

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