Block M = 7.50 kg is initially moving up the incline and is increasing speed with a = [09] m/s2 . The applied force F is horizontal. The coefficients of friction between the block and incline are μs = 0.443 and μk = 0.312. The angle of the incline is 25.0 degrees. a. What is the force F (N)? b. What is the normal force N (in N) between the block and incline? c. What is the force of friction (N) on the block?

a) 73.2N

b) 66.6N

c) 20.28N

Explanation:

F= mg=7.5×9.8=73.5N

Force parallel to the plane Fp= 73.5sin25= 31.06N

b) Fv= 73.5 cos25= 66.6N

c) Ff= u×Fv= 0.312×66.6=20.28N

a) Normal ForceFn= F/(cos25) - Fp - Ff = ma

1.1F -31.06-20.28=7.5×3.82

1.1F -51.84=28.65

1.1F= 28.65+51.84

F= 80.49/1.1

F= 73.2N

the temperature and the pressure affect solubility. temperature affects by changing the properties of the solvent. the increase in pressure increases solubility, the decrease in pressure decreases solubility.