### Problem StatementA particle is on a circular path as shown in the figure. At the position shown, the particle has a speed \( v = 5 \ \text{m/s} \), which is increasing at a rate of \( 7.5 \ \text{m/s}^2 \). What is the magnitude of its acceleration for the same position?### Diagram DescriptionThe diagram presents a circle with a radius of \( 2.5 \ \text{m} \). A particle is indicated by a blue dot on the circumference of the circle. The velocity vector (\( v \)) at the point where the particle is located is directed tangentially to the circle upwards, with \( v = 5 \ \text{m/s} \).### Choices for the Solution- \( a = 10 \ \text{m/s}^2 \)- \( a = 5 \ \text{m/s}^2 \)- \( a = 12.5 \ \text{m/s}^2 \)- \( a = 7.5 \ \text{m/s}^2 \)### ExplanationTo find the magnitude of the total acceleration \( a \), we need to consider both the tangential acceleration \( a_t \) and the centripetal (radial) acceleration \( a_r \).1. **Tangential Acceleration (\( a_t \)):** Given as \( 7.5 \ \text{m/s}^2 \).2. **Centripetal Acceleration (\( a_r \)):** This can be calculated using the formula: \[ a_r = \frac{v^2}{r} \] Where: - \( v = 5 \ \text{m/s} \) - \( r = 2.5 \ \text{m} \) Substituting the values: \[ a_r = \frac{(5)^2}{2.5} = \frac{25}{2.5} = 10 \ \text{m/s}^2 \]3. **Total Acceleration (\( a \)):** The magnitude of the total acceleration \( a \) is the vector sum of the tangential and centripetal accelerations: \[ a = \sqrt{a_t^2 + a_r^

Given a=7.5 m/s2 V=5 m/s r=2.5m

A particle is on a circular path as shown in the figure. At position shown, the particle has speed v = 5 m/s, which is increasing at a rate of 7.5 m/s². What is the magnitude of its acceleration for the same position? v = 5 m/s

A particle is on a circular path as shown in the figure. At position shown, the particle has speed v = 5 m/s, which is increasing at a rate of 7.5 m/s². What is the magnitude of its acceleration for the same position? v = 5 m/s

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Conservation of Energy

In physics, the rule of the conservation of energy states that energy doesn’t disappear from the atmosphere; instead, it changes its form from one type to another. Energy is one of the most important physical quantities in nature, and it cannot be created or destroyed — only transformed.

Question

### Problem Statement
A particle is on a circular path as shown in the figure. At the position shown, the particle has a speed \( v = 5 \ \text{m/s} \), which is increasing at a rate of \( 7.5 \ \text{m/s}^2 \). What is the magnitude of its acceleration for the same position?

### Diagram Description
The diagram presents a circle with a radius of \( 2.5 \ \text{m} \). A particle is indicated by a blue dot on the circumference of the circle. The velocity vector (\( v \)) at the point where the particle is located is directed tangentially to the circle upwards, with \( v = 5 \ \text{m/s} \).

### Choices for the Solution
- \( a = 10 \ \text{m/s}^2 \)
- \( a = 5 \ \text{m/s}^2 \)
- \( a = 12.5 \ \text{m/s}^2 \)
- \( a = 7.5 \ \text{m/s}^2 \)

### Explanation

To find the magnitude of the total acceleration \( a \), we need to consider both the tangential acceleration \( a_t \) and the centripetal (radial) acceleration \( a_r \).

1. **Tangential Acceleration (\( a_t \)):**
Given as \( 7.5 \ \text{m/s}^2 \).

2. **Centripetal Acceleration (\( a_r \)):**
This can be calculated using the formula:
\[
a_r = \frac{v^2}{r}
\]
Where:
- \( v = 5 \ \text{m/s} \)
- \( r = 2.5 \ \text{m} \)

Substituting the values:
\[
a_r = \frac{(5)^2}{2.5} = \frac{25}{2.5} = 10 \ \text{m/s}^2
\]

3. **Total Acceleration (\( a \)):**
The magnitude of the total acceleration \( a \) is the vector sum of the tangential and centripetal accelerations:
\[
a = \sqrt{a_t^2 + a_r^

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