What is the magnitude of the acceleration a of the chair? What is the magnitude of the normal force fn acting on the chair?

1. The acceleration of the chair is 0.64 m/s²

2. The normal force acting on the chair is 523.6 N

Explanation:

Let's list out the parameters given us:

mass = 45 kg, F = 144 N, θ = 35°, F (f) = 89 N

Solving for the horizontal component of the Force, we use the formula

F (hor) = F * cos θ = 144 * cos 35°

F (hor) = 117.96 N

F (hor) = 118 N

The friction force is acting upon the chair in the opposite direction of the f (hor) & must be accounted for

F (net) = F (hor) - F (f)

F (net) = 118 - 89 = 29

F (net) = 29N

F (net) = mg

g = F (net) ÷ m = 29 ÷ 45

g = 0.64 m/s²

F (norm) = F (chair) + F (ver)

F (chair) = mg

where g = acceleration due to gravity = 9.8 m/s²

F (chair) = 45 * 9.8 = 441 N

F (ver) = F * sin θ = 144 * sin 35°

F (ver) = 82.6 N

F (norm) = 441 + 82.6 = 523.6

F (norm) = 523.6 N

the results is a decrease in air temperature with an increase of altitude.

explanation:

volts times amps 6 x 0.5 3w