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ankoles [38]
3 years ago
11

A fixed end rectangular cantilever beam is subjected to 40 kg load at its end. The beam is 80 mm high, 20 mm wide and 0.5 m long

. What is the maximum deflection of the beam if the material used is high modulus graphite/epoxy composite (fiber volume fraction is
Physics
1 answer:
antoniya [11.8K]3 years ago
5 0

Answer:

The maximum deflection is 0.325 mm

Explanation:

we have the following data given:

m = 40 kg

L = 0.5 m

h = 80 mm = 0.08 m

b = 20 mm = 0.02 m

We calculate W:

W = m*g = 40 * 9.8 = 392.4 N

we will calculate the moment of inertia of the area which will be equal to:

I = (b*h^3)/12 = (0.02*0.08^3)/12 = 8.56x10^-7

The Young´s modulus is equal to:

E = (20*270)/(100) + (3.2*(100-20))/100 = 58.6x10^9 Pa

The deflection is equal to:

δ = (W*L^3)/(3*E*I) = (392.4 * (0.5^3))/(3*58.6x10^9*8.56x10^-7) = 0.000325 m = 0.325 mm

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Which of the following types of energy is potential energy A. gravitational energy
katovenus [111]

There are many forms of energy, but they can all be put into two categories: kinetic and potential. Kinetic energy is motion––of waves, electrons, atoms, molecules, substances, and objects. Potential energy is stored energy and the energy of position––gravitational energy

4 0
3 years ago
Following are the different layers of the sun’s atmosphere. Rank them based on the order in which a probe would encounter them w
blsea [12.9K]

Answer:

Going from earth to the sun a probe would encounter the next layers in order:

  • Corona
  • Transition Region
  • Chromosphere
  • Photosphere
  • Convection Zone
  • Radiative Zone
  • Core

A brief description of them:

Corona is the outermost layer and it cannot  be seen with the naked eye, is starts at about 2100 km from the surface of the sun and it has no limit defined.

Transition Region is between the corona and the chromosphere, it has an extension of about 100km

The chromosphere is between 400 km from the surface of the sun to 2100 km. In this layer the further you get away from the sun it gets hotter.

The photosphere is the surface of the sun, the part that we can see, and extends from the surface to 400km.

The convection zone is where convection happens, hot gas rises, cools and rises again.

Radiative Zone is where the photons try to rise to move to higher layers.

The core of the Sun is where nuclear fusion occurs due to the very high temperatures.

6 0
4 years ago
A child and sled with a combined mass of 58.0 kg slide down a frictionless slope. If the sled starts from rest and has a speed o
erma4kov [3.2K]

Answer:

The height is  h = 0.5224 \ m

Explanation:

From the question we are told that

   The combined mass of the child and the sled is  m =  58.0 \  kg

    The  speed of the sled is  u = 3.20 \ m/s

Generally applying SOHCAHTOA on the slope which the combined mass is down from

   Here the length of the slope(L)  where the combined mass slides through  is the hypotenuses

   while the height(h) of the height of the slope is the opposite

Hence from SOHCAHTOA

      sin (\theta) =  \frac{h}{L}

=>   Lsin(\theta) = h

Generally from the kinematic equation we have that

   v^2  = u^2 + 2aL

Here the u  is the initial velocity of the combined mass which is zero since it started from rest

 and  a is the acceleration of the combined mass which is mathematically evaluated as

       a =  g * sin (\theta )

      v^2  = u^2 + 2 *  g * sin (\theta )L

=>   2Lsin(\theta ) =  \frac{v^2 - u^2 }{g}

=>   h = \frac{ v^2 - u^2}{2g}

=>   h = \frac{ 3.20^2 - 0^2}{2 * 9.8 }

=>   h = 0.5224 \ m

6 0
3 years ago
A ship's propeller of diameter 3 m makes 10.6 revolutions in 30s. What is the angular velocity of the propeller?
ycow [4]

Answer:

The angular velocity of the propeller is 2.22 rad/s.

Explanation:

The angular velocity (ω) of the propeller is:  

\omega = \frac{\Delta \theta}{\Delta t}                              

Where:

θ: is the angular displacement = 10.6 revolutions

t: is the time = 30 s

\omega = \frac{\Delta \theta}{\Delta t} = \frac{10.6 rev*\frac{2\pi rad}{1 rev}}{30 s} = 2.22 rad/s

Therefore, the angular velocity of the propeller is 2.22 rad/s.

I hope it helps you!

5 0
3 years ago
A group of workers applied 10.0 networks of force to move a crate 20.0 meter. Calculate the work
olga2289 [7]

The work done is 200 J

Explanation:

The work done by a force applied to move an object is given by:

W= F d cos \theta

where

F is the magnitude of the force

d is the displacement of the object

\theta is the angle between the direction of the force and of the displacement

In this problem, assuming that the force applied by the workers is parallel to the direction of motion of the crate, we have:

F = 10.0 N

d = 20.0 m

\theta=0^{\circ}

Therefore, the work done is:

W=(10.0)(20.0)(cos 0)=200 J

Learn more about work here:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

4 0
4 years ago
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