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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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Answer:

3.62m/s and 2.83m/s

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Recall that

0.7813Vof = Vgf

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0.7880Vof + 0.4810Vof = 4.

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Vof = 3.62m/s

Recall that

0.7813Vof = Vgf

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3 0
3 years ago
What is the energy range (in joules) of photons of wavelength 410 nm to 750 nm ? Express your answers using two significant figu
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Answer:

4.9 x 10^-19 J, 2.7 x 10^-19 J

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Second wavelength, λ2 = 750 nm = 750 x 10^-9 m

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E = h c / λ

Where, h is the Plank's constant and c be the velocity of light.

h = 6.63 x 10^-34 Js

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So, energy correspond to first wavelength

E1 = (6.63 x 10^-34 x 3 x 10^8) / (410 x 10^-9) = 4.85 x 10^-19 J

E1 = 4.9 x 10^-19 J

So, energy correspond to second wavelength

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4 0
3 years ago
A 200kg ball on the end of string is swung in horizontal circle with radius of 0.5m . The ball makes revolution every 2second th
ANEK [815]

Answer:\dfrac{\pi}{2} ms^{-1}

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Let T be the time required to make one revolution.

Let r be the radius of the circular path.

Let d be the distance travelled by ball in one revolution.

As we know,the distance travelled in one revolution is the circumference of the circle.

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Speed of an object moving is circular path is define as the ratio of distance travelled in one revolution to the time taken by the object to complete one revolution.

Let s be the speed of the ball.

s=\frac{d}{T}=\frac{\pi }{2}ms^{-1}

So,the speed of the ball is \frac{\pi }{2}ms^{-1}

5 0
3 years ago
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