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Chapter C9, rolem011- Scogle Chrome Secure https/edugen.wieyplus.com/edugen/shared/assignment/test/qprint.uni Print by: MARCO GUZMAN PHYS 150 ONLINE JAN 2018 A/ Module 8 Problem Set (2016) Chapter 09, Problem 011 A big olve m = 0.063 kg lies at the origin or an xy coordinate system and a big Brazil ut M-0.35 kg lies at the point 0.79 1.5 m. Att = 0, a force 0 + j begins to act on the olive, and a force FN4.41 -2.0j|N begins to act on the nut. What is the (a) x and (b) y displacement of the center of mass of the olive-nut system et t 4.2 s, with respect to its position at t-o? Units (a) Number (b) Number Answer1: the number of signincant digits is set to 2; he toierance is +-I in the 2no signiicant dig Units Answer *2: the nunber of significant digits is set to 2: the tolerance is I in the 2nd significant digit Question Attempts: 0 of 10 used 11:05 AM Type here to scarch

Explanation / Answer

here,

for Olive

m = 0.083 kg

F1 = 2.8 i + 3.9 j N

accelration , a1 = F1/m

a1 = 33.7 i + 47 j m/s^2

at t1 = 4.2 s

the final position , r1 = (0 + 0.5 * a1x * t1^2 , 0.5 * a1y * t1^2 )

r1 = ( 297.2 , 414.6 ) m

for nut

M = 0.85 kg

F2 = - 4.4 i - 2 j N

accelration , a2 = F2/M

a2 = - 5.2 i - 2.35 j m/s^2

at t2 = 4.2 s

the final position , r2 = ( 0.79 + 0.5 * a2x * t2^2 , 1.5 + 0.5 * a2y * t2^2)

r2 = ( - 45.1 , - 19.2 ) m

the center of mass , r = (( m * x1 + M * x2 )/(m + M) , (m * y1 + M * y2 )/(m + M) )

r = ((0.083 * 297.2 - 0.85 * 45.1)/(0.083 + 0.85) , (0.083 * 414.6 - 0.85 * 19.2)/(0.083 + 0.85))

r = (- 14.7 , 19.4) m

a)

the x component of center of mass is - 14.7 m

b)

the y-component of center of mass is 19.4 m

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