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rewona [7]
3 years ago
14

72.0-kg person pushes on a small doorknob with a force of 5.00 N perpendicular to the surface of the door. The doorknob is locat

ed 0.800 m from axis of the frictionless hinges of the door. The door begins to rotate with an angular acceleration of 2.00 rad/s2. What is the moment of inertia of the door about the hinges?
Physics
1 answer:
expeople1 [14]3 years ago
4 0

Answer:I=2 kg-m^2

Explanation:

Given

mass m=72 kg

Force F=5 N

door knob is located at a distance of r=0.8 m from axis

Angular acceleration of door \alpha =2 rad/s^2

Torque T=I\alpha =F\times r

where I=moment of inertia

5\times 0.8=I\times 2

I=2 kg-m^2

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A small block of mass 20.0 grams is moving to the right on a horizontal frictionless surface with a speed of 0.68 m/s. The block
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Answer:

a) v'=-0.227\ m.s^{-1}

b) v=1.36\ m.s^{-1}

Explanation:

Given:

mass of the lighter block, m'=0.02\kg

velocity of the lighter block, u'=0.68\ m.s^{-1}

mass of the heavier block, m=0.04\ kg

velocity of the heavier block, u=0\ m.s^{-1}

a)

Using conservation of linear momentum:

m'.u'+m.u=m'.v'+m.v

where:

v'= final velocity of the lighter block

v= final velocity of the heavier block

m'.u'=m'.v'+m.v

m'(u'-v')=m.v ........................(1)

Since kinetic energy is conserved in elastic collision:

\frac{1}{2}m'.u'^2=\frac{1}{2}m'.v'^2+\frac{1}{2}m.v^2

m'(u'^2-v'^2)=m.v^2

m'(u'-v')(u'+v')= m.v^2

divide the above equation by eq. (1)

v=u'+v' .............................(2)

now we substitute the value of v from eq. (2) in eq. (1)

m'(u'-v')=m(u'+v')

\frac{m'+m}{m'-m} =\frac{u'}{v'}

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b)

now we substitute the value of v' from eq. (2) in eq. (1)

m'(u'-v+u')=m.v

2m'.u'=(m-m')v

2\times 0.02\times 0.68=(0.04-0.02)\times v

v=1.36\ m.s^{-1}

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