I NEED HELP PLEASE ASAP! :) A 2600-pound truck is stopped at a red light on a hill with an incline of 25°. Ignoring the force of friction, what force is required to keep the truck from rolling down the hill? (Show work)

( About ) 1,099 pounds

Step-by-step explanation:

The force required to keep this truck from rolling down the hill is opposed by the force of friction, comparative to the 2600 pounds of force against this friction. It should be that this opposing force is at work more towards the bottom of this hill, and thus the question asks the magnitude of latter force to keep this truck idle -

|| F || = sin( 25 ) * ( Fg ) = sin( 25 ) * ( 2600 lb ) = ( About ) 1098.80748 pounds

This can be rounded to a solution of 1099 pounds

intensity of the vector is v= √37 ≈ 6.08 units and make angle ∡α ≈ 9.46° with east direction

step-by-step explanation:

required vector is consists of the two componentsvx= 2+4=6 units   and vy= 1 unit and vx ⊥ vywe will use pythagorian theorem to find intensity of the vector vv∧2 = vx∧2 + vy∧2 => v = √vx∧2 + vy∧2 = √6∧2 + 1∧2 = √36+1 = √37 ≈ 6.08 unitsthe angle ∡α between vector and east direction we wil find with tanαtanα = 1/6 => α = arc tanα = arc 1/6 => α ≈ 9.46°

hope this