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Natalka [10]
3 years ago
8

A long uniform wooden board (a playground see-saw) has a pivot point at its center. An older child of mass M=32 kg is sitting a

distance L from the pivot. On the other side of the pivot point are two smaller children each of mass M/2. One is sitting a distance L/6 from the pivot. How far from the pivot must the other small child be sitting in order for the system to be balanced?
Physics
1 answer:
PIT_PIT [208]3 years ago
3 0

Answer:\frac{5L}{6}

Explanation:

Given

Wooden board is pivoted at center and

Older child of mass M=32\ kg is sitting at a distance of L from  center

if two child of mass \frac{M}{2} is sitting at a distance \frac{L}{6} and x(say) from pivot then net torque about pivot is zero

i.e.

\Rightarrow \tau_{net}=MgL-\frac{M}{2}g\frac{L}{6}-\frac{M}{2}gx

as \tau_{net}=0

Therefore

MgL=\frac{M}{2}g\frac{L}{6}+\frac{M}{2}gx

L-\frac{L}{6}=x

x=\frac{5L}{6}

Therefore another child is sitting at a distance of \frac{5L}{6}

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If a plane is flying level at 910 km/h and the banking angle is not to exceed 50 ∘, what's the minimum curvature radius for the
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Answer:

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Explanation:

First, we convert the distance from km/h to m/s

910 * 1000/3600

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making r the subject of the formula, we have

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4 0
3 years ago
*A car is going through a dip in the road whose curvature approximates a circle of radius 150m. At what velocity will the occupa
Valentin [98]

Answer:

v= 14.85 m/s

Explanation:

  • When at the bottom of the dip, the only force that keeps the car in the circular trajectory, is the centripetal force.
  • This force is not a new force, is just the net force aiming to the center of the circle.
  • In this case, is just the difference between the normal force (always perpendicular to the surface, pointing upward) and the force that gravity exerts on the car (which is known as the weight), pointing downward.
  • So, we can write the following expression:

       F_{cent} = F_{n} - F_{g}  (1)

  • It can be showed that the centripetal force is related to the speed by the following expression:
  • F_{cent} = m*\frac{v^{2}}{r} (2)
  • The normal force, it is called the apparent weight, because it would be the weight as measured by a scale.
  • Replacing (2) in (1), and solving for Fn, we get:

       F_{n} = m*\frac{v^{2} }{r} + m*g (3)

  • Now, we need to find the value of v that makes Fn, exactly 15% more than the weight m*g, so we can write the following equation:

      F_{n} = 1.15*F_{g} = m*\frac{v^{2}}{r} +F_{g}  (4)

  • Replacing Fg by its value, simplifying, and solving for v, we get:

       v = \sqrt{0.15*g*r} = \sqrt{9.8 m/s2*0.15*150m} = 14.85 m/s (5)

3 0
2 years ago
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