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Ierofanga [76]
1 year ago
14

The prop blades of an airplane spin with a linear velocity of 875 m/s and have a centripetal acceleration on the farthest edge o

f 180,000 m/s^2 the radius of the prop blades?
Physics
1 answer:
storchak [24]1 year ago
4 0

The radius of the prop blade of an airplane is determined as 4.25 m.

<h3>Radius of the prop blade</h3>

The radius of the prop blade of an airplane is calculated as follows;

a = v²/r

where;

  • v is the linear speed
  • r is the radius of the prop blade
  • a is the centripetal acceleration

r = v²/a

r = (875²)/(180,000)

r = 4.25 m

Thus, the radius of the prop blade of an airplane is determined as 4.25 m.

Learn more about centripetal acceleration here: brainly.com/question/79801

#SPJ1

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A string that passes over a pulley has a 0.341 kg mass attached to one end and a 0.625 kg mass attached to the other end. The pu
dalvyx [7]

Answer:

The frictional torque is \tau  = 0.2505 \ N \cdot m

Explanation:

From the question we are told that

   The mass attached to one end the string is m_1 =  0.341 \ kg

   The mass attached to the other end of the string is  m_2 =  0.625 \ kg

    The radius of the disk is  r = 9.00 \ cm  = 0.09 \ m

At equilibrium the tension on the string due to the first mass is mathematically represented as

      T_1 =  m_1 *  g

substituting values

      T_1 =  0.341 * 9.8

      T_1 =  3.342 \ N

At equilibrium the tension on the string due to the  mass is mathematically represented as

      T_2 =  m_2 *  g

     T_2 = 0.625 * 9.8

      T_2 = 6.125 \ N

The  frictional torque that must be exerted is mathematically represented as

      \tau  =  (T_2 * r ) - (T_1 * r )

substituting values  

     \tau  =  ( 6.125 * 0.09 ) - (3.342  * 0.09 )

     \tau  = 0.2505 \ N \cdot m

5 0
3 years ago
While traveling along a highway a driver slows from 24 m/sec to 15 m/sec in 12 seconds. What is the
Vladimir [108]
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4 0
3 years ago
A boat travels at 30 mph for a huge, solid cliff that is about 3,000 meters away. When the horn on the boat makes a toot, you ca
AleksandrR [38]

Answer:The frequency of the echo is slightly decreased

Explanation:

Given

speed of boat =30\ mph

cliff is 3000\ m away

when boat is still , suppose t is the time taken by the echo to reach observer on the boat

But as soon as boat starts moving  the distance between cliff and boat decreasing and time for echo to reach observer also decreases

and we know time \propto \frac{1}{frequency}

therefore frequency of the echo slightly decreased.

5 0
3 years ago
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