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igor_vitrenko [27]
3 years ago
11

The SI unit for force is the newton. Which of the following is equivalent to a newton? A. kg/m B. kg·m/s2 C. kg·m/s D. kg·m/s

Physics
2 answers:
bearhunter [10]3 years ago
8 0
I may be wrong but A
Bad White [126]3 years ago
6 0

<u>Answer:</u> The unit of force, newton is equivalent to kg.m/s^2

<u>Explanation:</u>

Force is defined as the push or pull on an object with some mass that causes change in its velocity.  It is also defined as the mass multiplied by the acceleration of the object. The S.I unit of force is Newton.

Mathematically,

F=m\times a

The S.I unit of mass is kilograms

The S.I unit of acceleration is m/s^2  (meters per second square)

So, the unit of force becomes: kg.m/s^2

Hence, the unit of force, newton is equivalent to kg.m/s^2

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An object is formed by attaching a uniform, thin rod with a mass of mr = 6.9 kg and length L = 4.88 m to a uniform sphere with m
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Answer:

a)  total moment of inertia is 1359.05 kg m^2

b) angular acceleratio is 0.854rad/sec^2

Explanation:

Given data:

m1=6.9 kg

L=4.88 m

m2=34.5 kg

R=1.22 m

we klnow that moment of inertia for rod is given as

J1=(1/12) ×m×L^2

J1 = (1/12) \times 6.9 \times 4.88^2 = 13.693 kg m^2

moment of inertia for sphere is given as

J1=(2/5) ×m×r^2

J1 = (2/5) \times 34.5 \times 1.22^2 = 20.539 kg m^2

As object rotates around free end of rod then for sphere the axis around what it rotates is at a distance of d2=L+R

For rod distance is  d1=0.5*L

By Steiner theorem

for the rod we get J_1'=J_1 + m_1\times d_1^2

J_1' = 13.693 + 2.44^2\times 6.9 = 54.77 kg m^2

for the sphere we get J_2' = J_2 + m_2\times d_2^2

J2' = 20.539 + 34.5\times 6.1^2 = 1304.28 kg^m2

And the total moment of inertia for the first case is

J_{t1} = J_1'+J_2' = 54.77 + 1304.28 = 1359.05kg.m^2

b) F=476 N

The torque for system is given as

M = F\times d\times sin(a)

where a is angle between Force and distance d

and where d represent distance from rotating axis.

In this case a = 90 degree  

M = F\times L/2

M=476*2.44 = 1161.44 Nm

The acceleration is calculated as

a_1 = \frac{M}{J_{t1}}

      = \frac{1161.44}{1359.05}

      = 0.854 rad/sec^2

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Answer:

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