1. a snowboarder of mass 60 kg is at rest on the top of a sand dune. the coefficients of static and kinetic friction are 0.55 and 0.4675, respectively. 60 kg μk = 0.4675 μs = 0.55 what is the angle the incline must exceed so that the snowboarder starts sliding? the acceleration of gravity is 9.8 m/s2 . answer in units of ◦. 2. what is the acceleration down the incline if = 35 ◦, exceeding the value found in the previous question? . i am so confused.

Inclined angle must exceed 26.57°

Acceleration = 1.17m/s^2

Explanation:

Mg= 60× 9.8=588N

Fp= force on inclined plain

Fp=588sinA

Fn=588cosA= normal force

Fs= us×Fn= 0.55(588cosA)= 323.4cosA

Fp-Fs=ma

588sinA - 323.4cosA=60×0

588sinA=323.4cosA

Dividing by 588

SinA=0.5cosA

But sinA/cosA = Tan

Tan A=0.5

Tan^-0.5 = 26.57°

Weight(x) - Fn= ma

Sin35mg - ucos 35mg = ma

Dividing through by m gives

Sin35g- ucosg= a

0.5736×9.8)- (0.55×0.8192×9.8) =a

5.62- 4.42=a = 1.17m/s^2

if the thermos has a vacuum innner, then heat through conv currents and conduction are reduced = vacuum flask

the answer would be c. motion and direction

i hope this : )