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 m/s2 direction ° (measured clockwise from the vertical)

The acceleration will be 18 m/s², 6° from the vertical.

Explanation:

The resulting force will be the sum of both forces (see figure):

F flea = (0, 1.07 × 10⁻⁵ N)

F wind = (1.12 × 10⁻⁶ N, 0)

Resulting force = F flea + F wind

R F = (0, 1.07 × 10⁻⁵ N) +  (1.12 × 10⁻⁶ N, 0) = ( 1.12 × 10⁻⁶ N, 1.07 × 10⁻⁵ N)

The magnitude of the resulting force will be:

The magnitude of the acceleration will be:

F = m·a

Where

F = force

m = mass

a = acceleration

F/m = a

1.08 × 10 ⁻⁵N / 6.0 × 10⁻⁷ kg = 18 m/s²

The direction of the acceleration will be the same as the direction of the resulting force. Using trigonometry:

cos α = adjacent/hypotenuse = 1.12 × 10⁻⁶ N / 1.08 × 10⁻⁵ N

α = 84°

Then the acceleration will be:

18 m/s² (90° - 84°) 6° from the vertical.

wavelength divides in half and speed stays the same.