Hello I need help solving those questions please

9! d in the diagram shown below, the lower block is acted on by a force, F, which has a magnitude of 94.0N. The coefficient of kinetic friction between the lower block and the surface is 0.339. The coefcient of kinetic friction between the lower block and the upper block is also 0.339. What is the acceleration of the lower block, if the mass of the lower block is 4.84kg and the mass of the upper block is 1.67kg? (8) A block of mass M1 resting on a 21.60 slope is shown. The block has coefficients of friction Us = 0.724 and HK = 0.422 with the surface. It is connected via a massless string over a massless, frictionless pulley to a hanging block of mass M2 = 2.98 kg. M1 M2 (8a) What is the minimum mass M1 that will remain stationary and not slip? Submit Answer Tries 0/10 (8b) If this minimum mass is nudged ever so slightly, it will start being pulled up the incline. What magnitude of acceleration will the minimum mass have?(8a) What is the minimum mass M1 that will remain stationary and not slip? Submit Answer Tries 0/10 (8b) If this minimum mass is nudged ever so slightly, it will start being pulled up the incline. What magnitude of acceleration will the minimum mass have? Submit Answer Tries 0/10 The 1.03kg physics book shown is connected by a string to a 502.0g coffee cup. The book is given a push up the slope (call this direction positive) and released with a speed of 4.39m/s. The coefficients of friction are Us = 0.557 and HK = 0.110. What is the acceleration of the book if the slope is inclined at 26.8? Submit Answer Tries 0/10 How far does the book slide before it has zero speed?In the drawing, the rope and the pulleys are massless, and there is no friction. The larger block has a mass of 9.92kg and the smaller block has a mass of 3.84kg. Find the tension in the rope. (Hint: the larger mass moves twice as far as the smaller mass). Submit Answer Tries 0/10 What is the acceleration of the 9.92kg block?A 1.20kg ball is connected by means of two massless strings to a vertical, rotating rod. The strings are tied to the rod and form two sides of an equilateral triangle. The tension in the upper string is 56.8N. String length - r Rotating rod If the length of each string is 1.72m, what is the tension in the lower string? Submit Answer Tries 0/10 What is the speed of the ball?( 3) Cup and Book ( as Here , we forget to write the j' term in the final equation of acceleration , re Q = ( m , + Ulx my Cost + m , sing ) g ( m , f my ) 7 a = - (0 . 7 ) x 9 .8 a = - 6. 86 m /s 2 la = 6.86 m/s ( down the incline (b) Using relation, V u = 2as 0 - ( 4 3 9 ) = 2 X ( - 6. 86 ) XS S = 1. 4 m -s up the incline .( 2 ) we have not taken the friction of mass m, on mass my . So, Free body diagram of my will be, T F f 2 m N 2 Sfy = 0 > F- T - f,- f2 = mza 2 ) F -T - UIN , - UN =ma - ( 2 1 F - T - MIN, - U. ( N , + M 2 g ) = m za J F - T - 1, mig - el2 might liz mig = mizq = ) F - T - ( el , m , + 412m , + 12 m z ) g = my a Put the value of 'T' from ea"( 1) of previous solution. 2 ) F - ( mate , m ,g ) - ( Mim , + 1/2 m , + 1, my ) 9 = m z g F - 2 1 , mig + 12 ( m it m z ) g = m z 9 + m , a a = F - [ 2 4 , m , + 1 2 ( m , + m, ) ] g ( mit m . ) But the values , we get - a = 94 - | 2 x 0 . 3 3 9 x 1 - 6 7 + 0 . 33 9 ( 1 . 67 + 4. 84 ) x 9.8 ( 1.67 + 4. 84 ) a = 9 . 41 m / s 2