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ankoles [38]
2 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]2 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: The observing friend will the swimmer moving at a speed of 0.25 m/s.

Explanation:

  • Let <em>S</em> be the speed of the swimmer, given as 1.25 m/s
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  • Note that this speed is the magnitude of the velocity which is a vector quantity.
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Hence the resultant velocity is given as, S_{R} = S — S 0S_{0}

S_{R} = 1.25 — 1

S_{R} = 0.25 m/s.

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A very long uniform line of charge has charge per unit length λ1 = 4.68 μC/m and lies along the x-axis. A second long uniform li
Kitty [74]

Answer:

E_{net} = 6.44 \times 10^5 N/C

Explanation:

As we know that electric field due to infinite line charge distribution at some distance from it is given as

E = \frac{2k \lambda}{r}

now we need to find the electric field at mid point of two wires

So here we need to add the field due to two wires as they are oppositely charged

Now we will have

E_{net} = \frac{2k\lambda_1}{r} + \frac{2k\lambda_2}{r}

now plug in all data

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\lambda_2 = 2.48 \mu C/m

r = 0.200 m

now we have

E_{net} = \frac{2k}{r}(4.68 + 2.48)

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E_{net} = 6.44 \times 10^5 N/C

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