### Problem DescriptionA flea jumps by exerting a force of \( 1.02 \times 10^{-5} \) N straight down on the ground. A breeze blowing on the flea, parallel to the ground, exerts a force of \( 1.20 \times 10^{-6} \) N on the flea. Find the direction and magnitude (in m/s²) of the acceleration of the flea if its mass is \( 6.0 \times 10^{-7} \) kg. (Let us assume that \( \vec{F}_{\text{wind}} \) points to the right. We will consider this to be the +x direction and vertical to be the +y direction.)#### Provided Values:- Force exerted by the flea: \( 1.02 \times 10^{-5} \) N (downward, -y direction)- Force due to breeze: \( 1.20 \times 10^{-6} \) N (rightward, +x direction)- Mass of the flea: \( 6.0 \times 10^{-7} \) kg#### Required:Calculate the following for the flea's acceleration:- Magnitude \( (\text{in } \text{m/s}^2) \)- Direction \( (\text{in degrees measured clockwise from the vertical}) \)### Solution Steps1. **Calculate the Acceleration Components:** Given: \[ \vec{F}_{\text{flea}} = -1.02 \times 10^{-5} \, \text{N} \quad (\text{downward, \(-y\) direction}) \] \[ \vec{F}_{\text{wind}} = 1.20 \times 10^{-6} \, \text{N} \quad (\text{rightward, \(+x\) direction}) \] Mass \( m = 6.0 \times 10^{-7} \, \text{kg} \) Using Newton's Second Law, \( \vec{a} = \frac{\vec{F}}{m} \), we calculate the acceleration components: \[ a_x = \frac{\vec{F}_{\text{wind}}}{m} = \frac{1.20 \times 10^{-6}}{6

Data Given , Force straight down on the ground ( F1 ) = 1.02 × 10-5 N Force parallel to the ground (…

A flea jumps by exerting a force of 1.02 x 10 5N straight down on the ground. A breeze blowing on the flea parallel to the ground exerts a force of 1.20 x 10 6 N on the flea. Find the direction and magnitude (in m/s²) of the acceleration of the flea if its mass is 6.0 x 10 kg. (Let us assume that Fwind points to the right. We will consider this to be the +x direction and vertical to be the +y direction.) magnitude m/s2 direction ° (measured clockwise from the vertical)

A flea jumps by exerting a force of 1.02 x 10 5N straight down on the ground. A breeze blowing on the flea parallel to the ground exerts a force of 1.20 x 10 6 N on the flea. Find the direction and magnitude (in m/s²) of the acceleration of the flea if its mass is 6.0 x 10 kg. (Let us assume that Fwind points to the right. We will consider this to be the +x direction and vertical to be the +y direction.) magnitude m/s2 direction ° (measured clockwise from the vertical)

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter6: Energy Of A System

Section: Chapter Questions

Problem 67P

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Question

### Problem Description

A flea jumps by exerting a force of \( 1.02 \times 10^{-5} \) N straight down on the ground. A breeze blowing on the flea, parallel to the ground, exerts a force of \( 1.20 \times 10^{-6} \) N on the flea. Find the direction and magnitude (in m/s²) of the acceleration of the flea if its mass is \( 6.0 \times 10^{-7} \) kg. (Let us assume that \( \vec{F}_{\text{wind}} \) points to the right. We will consider this to be the +x direction and vertical to be the +y direction.)

#### Provided Values:
- Force exerted by the flea: \( 1.02 \times 10^{-5} \) N (downward, -y direction)
- Force due to breeze: \( 1.20 \times 10^{-6} \) N (rightward, +x direction)
- Mass of the flea: \( 6.0 \times 10^{-7} \) kg

#### Required:
Calculate the following for the flea's acceleration:
- Magnitude \( (\text{in } \text{m/s}^2) \)
- Direction \( (\text{in degrees measured clockwise from the vertical}) \)

### Solution Steps

1. **Calculate the Acceleration Components:**

Given:
\[
\vec{F}_{\text{flea}} = -1.02 \times 10^{-5} \, \text{N} \quad (\text{downward, \(-y\) direction})
\]
\[
\vec{F}_{\text{wind}} = 1.20 \times 10^{-6} \, \text{N} \quad (\text{rightward, \(+x\) direction})
\]

Mass \( m = 6.0 \times 10^{-7} \, \text{kg} \)

Using Newton's Second Law, \( \vec{a} = \frac{\vec{F}}{m} \), we calculate the acceleration components:
\[
a_x = \frac{\vec{F}_{\text{wind}}}{m} = \frac{1.20 \times 10^{-6}}{6

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