Aflea jumps by exerting a force of 1.07 10-5 n straight down on the ground. a breeze blowing on the flea parallel to the ground exerts a force of 1.12 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 10-7 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 9.98 m/s2 direction ° (measured clockwise from the vertical)

Explanation:

A flea jumps by exerting a force of 1.07 10⁻⁵ N straight down on the ground. It will receive 1.07 x10⁻⁵ N as reaction force which will create an acceleration of a in it so that

1.07x 10⁻⁵ - mg = ma

1.07x 10⁻⁵ - 6 x 10⁻⁷ x 9.8  = 6 x 10⁻⁷  a

48.2  x 10⁻⁷ =  6 x 10⁻⁷  a

a_y = 8 m / s

acceleration in horizontal direction

a_x = 1.12 x 10⁻⁶ / 6 x 10⁻⁷

a_x = 1.12 x 10⁻⁶ / 0.6 x 10⁻⁶

= 1.87 m /s

resultant acceleration

= √(a_x² +a_y²)

= √(1.87² + 8²)

= 8.21 m / s

If angle be θ clockwise from vertical

Tanθ = 1.87 / 8

θ = 13 degree.

well clustered together.

points close on a scatter diagram