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Nadusha1986 [10]
3 years ago
15

A child of mass 47 kg sits on the edge of a merry-go-round with radius 1.3 m and moment of inertia 56.3953 kg m2 . The merrygo-r

ound rotates with an angular velocity of 1.6 rad/s. What radial force does the child have to exert to stay on the merry-go-round?
Physics
1 answer:
lakkis [162]3 years ago
4 0

Answer:

F=156.416\ N the child have to exert this much amount of force radially to stay on the wheel.

Explanation:

Given:

mass of the child, m=47\ kg

radius of the merry-go round wheel, r=1.3\ m

moment of inertia of the wheel, I=56.3953\ kg.m^2

angular velocity of the wheel, \omega=1.6\ rad.s^{-1}

Since the child is sitting on the edge of the rotating wheel the child will feel and outward throwing force called centrifugal force.

Centrifugal force is mathematically given as:

F=m.r.\omega^2

F=47\times 1.3\times 1.6^2

F=156.416\ N the child have to exert this much amount of force radially to stay on the wheel.

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an 1150kg elevator moving down speeds up at a rate of 3.5m/s. what is the tension in the supporting cables?
gtnhenbr [62]

Answer:

The tension force in the supporting cables is 7245N

Explanation:

There are two forces acting on the elevator: the force of gravity pointing down (+) with magnitude (elevator mass) x (gravitational acceleration), and the tension force of the cable pointing up (-) with an unknown magnitude F. The net force is the sum of these forces:

F_{net} = F_g - F = m\cdot g - F\\

We are given the resulting acceleration along with the mass, i.e., we know the net force, allowing us to solve for F:

1150kg\cdot 3.5\frac{m}{s^2}= 1150kg \cdot 9.8\frac{m}{s^2}-F\\\implies F = 1150kg\cdot(9.8-3.5)\frac{m}{s^2}= 7245N

The tension force F in the supporting cables is 7245N


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A 4 kg toy car moves horizontally on a rough road with coefficient of kinetic friction 0.2. It accelerates from rest to 20 m/s i
attashe74 [19]

The total work done on the car is 784Joule.

<h3>What's the acceleration of the car?</h3>
  • As per Newton's equation of motion, V= U+at
  • U= initial velocity= 0 m/s

V= vinal velocity= 20m/s

t= time = 10s

a= acceleration

  • So, 20= 0+ 10a

=> a= 20/10= 2m/s²

<h3>What's the distance covered by the car in 10 seconds?</h3>
  • As per Newton's equation of motion,

V²-U² = 2aS

  • S= distance covered by the car
  • So, 20²-0=2×2×S=4S

=> 400= 4S

=> S= 400/4= 100m

<h3>What's the work done on the car due to frictional force?</h3>

Work done by frictional force= frictional force × distance

= (0.2×4×9.8)×100

= 784Joule

Thus, we can conclude that the work done on the car is 784Joule.

Learn more about the work done here:

brainly.com/question/25573309

#SPJ1

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