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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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zysi [14]

As per the question, the mass of meteorite [ m]= 50 kg

                       The velocity of the meteorite [v] = 1000 m/s

When the meteorite falls on the ground, it will give whole of its kinetic energy to earth.

We are asked to calculate the gain in kinetic energy of earth.

The kinetic energy of meteorite is calculated as -

                                       Kinetic\ energy\ [K.E]\ =\frac{1}{2} mv^2

                                                             =\frac{1}{2}50kg*[1000\ m/s]^2

                                                               =\frac{1}{2}50* 10^{6}\ J

                                                               =25*10^6\ J    

Here, J stands for Joule which is the S.I unit of energy.

Hence,\ the\ kinetic\ energy\ gained\ by\ earth\ is\ 25*10^6\ J

4 0
3 years ago
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You are standing on a sheet of ice that covers the football stadium parking lot in Buffalo; there is negligible friction between
Anni [7]

Answer:

0.074m/s

Explanation:

We need the formula for conservation of momentum in a collision, this equation is given by,

m_1u_1+m_2u_2 = m_1v_1+m_2v_2

Where,

m_1 = mass of ball

m_2 = mass of the person

u_1 = Velocity of ball before collision

u_2 = Velocity of the person before collision

v_1 = velocity of ball afer collision

v_2= velocity of the person after collision

We know that after the collision, as the person as the ball have both the same velocity, then,

v_1 = v_2

m_1u_1 + m_2u_2 = (m_1+m_2)v_2

Re-arrenge to find v_2,

v_2 = \frac{m_1u_1+m_2u_2}{m_1+m_2}

Our values are,

m_1= 0.425kg

u_1= 12m/s

m_2= 68.5kg

u_2= 0m/s

Substituting,

v_2 = \frac{(0.425)(12)+(68.5)(0)}{0.425+68.5}

v_2 = 0.074m/s

<em />

<em>The speed of the person would be 0.074m/s after the collision between him/her and the ball</em>

7 0
3 years ago
On the sonometer shown below, a horizontal cord of length 5 m has a mass of 1.45 g. When the cord was plucked the wave produced
Korolek [52]

Answer:

(a) T = 0.015 N

(b) M = 1.53 x 10⁻³ kg = 1.53 g

Explanation:

(a) T = 0.015 N

First, we will find the speed of waves:

v =f\lambda

where,

v = speed of wave = ?

f = frequency = 120 Hz

λ = wavelength = 6 cm = 0.06 m

Therefore,

v = (120 Hz)(0.06 m)

v = 7.2 m/s

Now, we will find the linear mass density of the coil:

\mu = \frac{m}{l}

where,

μ = linear mass density = ?

m = mass = 1.45 g = 1.45 x 10⁻³ kg

l = length = 5 m

Thereforre,

\mu = \frac{1.45\ x\ 10^{-3}\ kg}{5\ m}\\\\\mu = 2.9\ x\ 10^{-4}\ kg/m

Now, for the tension we use the formula:

v = \sqrt{\frac{T}{\mu}}\\\\7.2\ m/s = \sqrt{\frac{T}{2.9\ x\ 10^{-4}\ kg/m}}\\\\(51.84\ m^2/s^2)(2.9\ x\ 10^{-4}\ kg/m) = T

<u>T = 0.015 N</u>

<u></u>

(b)

The mass to be hung is:

T = Mg\\\\M = \frac{T}{g}\\\\M = \frac{0.015\ N}{9.8\ m/s^2}\\\\

<u>M = 1.53 x 10⁻³ kg = 1.53 g</u>

4 0
3 years ago
Which symbol and unit of measurement are used for electric current?
Burka [1]

Answer: Symbol is I and unit A

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4 0
3 years ago
A 5,000 kg satellite is orbiting the Earth in a circular path. The height of the satellite above the surface of the Earth is 800
Arada [10]

Explanation:

The given data is as follows.

            m = 5000 kg,            h = 800 km = 0.8 \times 10^{6} m

    R_{e} = 6.37 \times 10^{6} m,    r = R + h = 7.17 \times 10^{6} m

   M_{e} = 5.98 \times 10^{24} kg,   G = 6.67 \times 10^{-11} Nm^{2}/kg^{2}

As we know that,

              \frac{mv^{2}}{r} = \frac{GmM_{e}}{r^{2}}

                      v = \sqrt{\frac{GM_{e}}{r^{2}}}

And, it is known that formula to calculate angular velocity is as follows.

               \omega = \frac{v}{r} = \sqrt{\frac{GM_{e}}{r^{3}}}

                            v = \sqrt{\frac{GM_{e}}{r^{3}}}

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                                = 1.0402 \times 10^{-3} rad/s

Thus, we can conclude that speed of the satellite is 1.0402 \times 10^{-3} rad/s.

6 0
3 years ago
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