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.

    m₁v₁ + m₂v₂ = (m₁ + m₂)vmomentum is conserved.the left-hand side of the equation represents the momentum in the system before the collision.the right-hand side represents the momentum in the system after the collisionm₁ and v₁ represents the mass and the velocity of the physics blockm₂ and v₂ represents the mass and velocity of the ball of puttyv represents the final velocity of the combined mass.the question states the two masses stick together. they combined mass has its own velocity with the body mass the sum of the two objects' massestheir combined momentum is (m₁ + m₂)vto find the magitude of (m₁ + m₂)v, we need to add the vectors on the left-hand side of the equation: m₁v₁ + m₂v₂.note the diagram. i have positioned the vectors in accordance with their directions.the magnitude of the resultant can be calculated using the pythagorean theorem since this is a right triangle situation (north and west are perpendicular)|(m₁ + m₂)v| =  √[ (m₁v₁)² + (m₂v₂)² ]|(m₁ + m₂)v| =  √[ (1.5kg·8.0m/s)² + (0.40kg·20m/s)² ]|(m₁ + m₂)v| =  14.4 kg m/sfor two significant figures (assuming that 20m/s is two), this rounds to 14 kg m/s