/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 75 A marble is released from one ri... [FREE SOLUTION] | 91Ó°ÊÓ

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A marble is released from one rim of a hemispherical bowl of diameter 50.0 \(\mathrm{cm}\) and rolls down and up to the opposite rim in 10.0 s. Find (a) the average speed and (b) the average velocity of the marble.

Short Answer

Expert verified
(a) Average speed: 0.07854 m/s. (b) Average velocity: 0.05 m/s.

Step by step solution

01

Understand the Problem

A marble moves in a hemispherical bowl with a diameter of 50.0 cm. We need to find both the average speed and average velocity as it travels from one rim to the opposite rim in 10.0 seconds.
02

Calculate the Distance for Average Speed

Since the marble travels from one rim to the opposite rim along the circular path (half a circle), calculate the circumference of the full circle. Use the formula for circumference: \(C = \pi \times d\), where \(d\) is the diameter of the circle.
03

Apply the Formula for Circumference

Substitute the given diameter into the circumference formula: \[ C = \pi \times 50.0 \text{ cm} = 157.08 \text{ cm} \]
04

Determine Half Circumference

Since the marble travels half the circumference, we divide by 2: \[ \text{Distance} = \frac{157.08}{2} = 78.54 \text{ cm} = 0.7854 \text{ m} \] (convert cm to meters by dividing by 100)
05

Compute the Average Speed

Average speed is calculated as: \( \text{Average speed} = \frac{\text{Total distance}}{\text{Time}} \). Substitute the distance and time: \[ \text{Average speed} = \frac{0.7854}{10} = 0.07854 \text{ m/s} \]
06

Determine Net Displacement for Average Velocity

Displacement is the straight line distance between the starting and ending points. Here, it is equal to the diameter of the bowl: 50.0 cm or 0.5 m.
07

Compute the Average Velocity

Average velocity is the total displacement divided by the total time. Substitute the displacement and time: \[ \text{Average velocity} = \frac{0.5}{10} = 0.05 \text{ m/s} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Average Speed
Average speed is an essential concept in kinematics, representing the total distance traveled divided by the total time taken. It provides a sense of how fast an object moves regardless of its direction. In our example, the marble rolls along half the circumference of a hemispherical bowl. To calculate average speed, we need to determine this path's distance and divide it by the time the marble takes to complete it.
The marble follows a path that's half the circle's circumference because it travels from one rim of the bowl to the other. First, we compute the full circumference using the formula for a circle's circumference:
  • Calculate the full circle's circumference: \( C = \pi \times d \)
  • For our bowl, the diameter \( d \) is 50 cm, so \( C = \pi \times 50 \text{ cm} = 157.08 \text{ cm} \).
  • The marble travels half this distance: \( 78.54 \text{ cm} \).
  • Converting to meters gives \( 0.7854 \text{ m} \).
If it takes 10 seconds for the marble to travel this distance, we use the formula for average speed:
  • \( \text{Average speed} = \frac{\text{Total distance}}{\text{Time}} = \frac{0.7854}{10} = 0.07854 \text{ m/s} \).
This calculation helps us understand the marble's overall pace as it moves along its curving path.
Average Velocity
While average speed focuses on the total distance covered, average velocity considers the total displacement in a specific direction. Displacement measures the straight-line distance from the start to finish, always considering direction, which is crucial when paths are not straight.
In the marble's case, displacement is simply the diameter of the hemispherical bowl. Unlike average speed, displacement doesn't factor in the curving path but rather the linear distance between start and end points. Here, it translates to 50 cm, or 0.5 m.
To ascertain the average velocity, we apply:
  • \( \text{Average velocity} = \frac{\text{Displacement}}{\text{Time}} = \frac{0.5}{10} = 0.05 \text{ m/s} \).
This result shows us how fast and in what direction the marble is displacing across its bowl journey, demonstrating the importance of both magnitude and direction in velocity calculations.
Displacement
Displacement is a fundamental concept in kinematics, reflecting the shortest distance between an object's starting and ending positions, taking direction into account. Unlike distance, displacement is a vector quantity, meaning it embodies both magnitude and direction.
In our scenario with the marble, displacement is calculated as the straight line that connects the two rim points of the hemispherical bowl. This path is the bowl's diameter, which simplifies the displacement distance to 50 cm, or emblematically converting scales to 0.5 m.
  • It's important to note that the trabvel path (curvature) is not considered in displacement, only the net change in position.
Understanding displacement helps distinguish between covering ground and actual movement concerning direction, enhancing one's grasp of motion dynamics in practical symmetry-laden situations.
Circumference Calculation
Circumference calculation plays a pivotal role in determining path length for circular motions. It is key for assessing distances in systems like circular tracks, or in this case, the hemispherical bowl.To find the circumference of a circle, apply the formula \( C = \pi \times d \), where \( d \) is the diameter. In our exercise, the entire circumference calculation helps address the marble's travel distance. Given:
  • Diameter of the bowl: 50 cm
  • Full circle's circumference: \( C = \pi \times 50 \text{ cm} = 157.08 \text{ cm} \)
  • Since the trip is half the circle, the effective travel is \( \frac{157.08}{2} = 78.54 \text{ cm} \)
  • Converting for practical use, \( 0.7854 \text{ m} \)
Recognizing circumference in partial travels, such as half of it, allows better understanding of real travel distances in circular motions, essential for calculating average speeds and orbital-like movements.

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Most popular questions from this chapter

Starting from the front door of your ranch house, you walk 60.0 \(\mathrm{m}\) due east to your windmill, and then you turn around and slowly walk 40.0 \(\mathrm{m}\) west to a bench where you sit and watch the sunrise. It takes you 28.0 s to walk from your house to the windmill and then 36.0 s to walk from the windmill to the bench. For the entire trip from your front door to the bench, what are (a) your average velocity and (b) your average speed?

A Multistage Rocket. In the first stage of a two-stage rocket, the rocket is fired from the launch pad starting from rest but with a constant acceleration of 3.50 \(\mathrm{m} / \mathrm{s}^{2}\) upward. At 25.0 s after launch, the second stage fires for \(10.0 \mathrm{s},\) which boosts the rocket's velocity to 132.5 \(\mathrm{m} / \mathrm{s}\) upward at 35.0 s after launch. This firing uses up all the fuel, however, so after the second stage has finished firing, the only force acting on the rocket is gravity. Air resistance can be neglected. (a) Find the maximum height that the stage-two rocket reaches above the launch pad. (b) How much time after the end of the stage-two firing will it take for the rocket to fall back to the launch pad? (c) How fast will the stage-two rocket be moving just as it reaches the launch pad?

An astronaut has left the International Space Station to test a new space scooter. Her partner measures the following velocity changes, each taking place in a 10 -s interval. What are the magnitude, the algebraic sign, and the direction of the average acceleration in each interval? Assume that the position is to the right. (a) At the beginning of the interval the astronaut is moving toward the right along the \(x\) -axis at \(15.0 \mathrm{m} / \mathrm{s},\) and at the end of the interval she is moving toward the right at 5.0 \(\mathrm{m} / \mathrm{s} .\) (b) At the beginning she is moving toward the left at \(5.0 \mathrm{m} / \mathrm{s},\) and at the end she is moving toward the left at 15.0 \(\mathrm{m} / \mathrm{s}\) (c) At the beginning she is moving toward the right at \(15.0 \mathrm{m} / \mathrm{s},\) and at the end she is moving toward the left at 15.0 \(\mathrm{m} / \mathrm{s}\).

A rocket carrying a satellite is accelerating straight up from the earth's surface. At 1.15 s after liftoff, the rocket clears the top of its launch platform, 63 \(\mathrm{m}\) above the ground. After an additional \(4.75 \mathrm{s},\) it is 1.00 \(\mathrm{km}\) above the ground. Calculate the magnitude of the average velocity of the rocket for (a) the 4.75 -s part of its flight and (b) the first 5.90 s of its flight.

An entertainer juggles balls while doing other activities. In one act, she throws a ball vertically upward, and while it is in the air, she runs to and from a table 5.50 \(\mathrm{m}\) away at a constant speed of 2.50 \(\mathrm{m} / \mathrm{s}\) , returning just in time to catch the falling ball. (a) With what minimum initial speed must she throw the ball upward to accomplish this feat? (b) How high above its initial position is the ball just as she reaches the table?

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