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Annette [7]
1 year ago
13

in the primitive yo-yo apparatus (figure 1), you replace the solid cylinder with a hollow cylinder of mass m , outer radius r ,

and inner radius r/2 . find the magnitude of the downward acceleration of the hollow cylinder.
Physics
1 answer:
kirza4 [7]1 year ago
4 0

The magnitude of the downward acceleration of the hollow cylinder is 6m/s^2.

Z = I α

T.R =1/2 M ( R^{2} + (R/2)^{2} )α

T.R = 1/2M 5R^{2}/4 α

T = 5Ma/8

Mg - T = Ma

Mg -  5Ma/8 =  Ma

Mg= 5Ma/8 +  Ma = 13Ma / 8

acceleration = 8g/13 = 6 m/s^2

The rate at which an object's velocity with respect to time changes is called its acceleration. The direction of the net force imposed on an item determines its acceleration in relation to that force. According to Newton's Second Law, the magnitude of an object's acceleration is the result of two factors working together

The size of the net balance of all external forces acting on that item is directly proportional to the magnitude of this net resultant force; the magnitude of that object's mass, depending on the materials from which it is built, is inversely related to its mass.

Learn more about acceleration here:

brainly.com/question/2303856

#SPJ4

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At time 10.28 s after A is fired bullet B passes A.

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Explanation:

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s=ut+0.5at^2 \\

Here s = h,u = 450m/s a = -g and t = t+3

Substituting

h=450(t+3)-0.5\times 9.81\times (t+3)^2=450t+1350-4.9t^2-29.4t-44.1\\\\h=420.6t-4.9t^2+1305.9

Distance traveled by second bullet

Here s = h,u = 600m/s a = -g and t = t

Substituting

h=600t-0.5\times 9.81\times t^2=600t-4.9t^2\\\\h=600t-4.9t^2 \\

Solving both equations

600t-4.9t^2=420.6t-4.9t^2+1305.9\\\\179.4t=1305.9\\\\t=7.28s \\

So at time 10.28 s after A is fired bullet B passes A.

Height at t = 7.28 s

h=600\times 7.28-4.9\times 7.28^2\\\\h=4108.31m \\

Passing of B occurs at 4108.31 height.

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