Posner, Snyder, and Davidson (1980)examined spatial attention using arrows as primes. In most of the trials the arrow pointed to the area where the stimulus would appear a moment later, but 20% of the time it did not. They compared reaction times (RTs)for trials when the cue correctly pointed to the stimulus location, for trials when the arrow pointed toward an incorrect position, and for trials with a neutral cue (no arrow).

Which of the following statements was NOT supported by their findings?

A)RTs were slower when the arrow pointed toward an incorrect location.

B)Spatial attention is limited in capacity.

C)We can attend to two different locations without a reduction in performance.

D)RTs were faster for responses for arrows pointing correctly compared to the neutral condition.

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    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