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VARVARA [1.3K]
2 years ago
10

A water tower is idealized as a mass M on top of a uniform and massless beam. The bottom end of the beam is fixed to the ground.

The beam has solid circular cross section with a diameter of 1.2 m. Its Young’s modulus is 30 GPa, and its length is 10 m. The mass M is 20 tons. Find the natural frequency and natural period of this system in lateral oscillations.
Physics
1 answer:
Tems11 [23]2 years ago
8 0

Answer:

Natural frequency=21.40 Hz

Time= 0.2936 seconds

Explanation:

Idealizing the question as a cantilever beam with point load of mass M as 20 tons

Lateral stiffness, k=\frac {3EI}{l^{3}} where l is length given as 10 m, E is Young’s modulus given as 30GPa and I is inertia where for a circular cross-section is given by \frac {\pi d^{4}}{64}

k=\frac {3*(30*10^{9})*(\pi *1.2^{4})}{64*10^{3}}= 9160884.178

k= 9.160884178*10^{6}

To find the frequency, w_{n}, the mass m is given as 20 tons or 20000 Kg

w_{n}=\sqrt (\frac {k}{m})= \sqrt (\frac {9.160884178*10^{6}}{20000})=21.40196741 Hz

Natural frequency=21.40 Hz

Time period,

T=\frac {2\pi}{w_{n}}=\frac {2\pi}{21.40196741}=0.2935798 seconds

T=0.2936 seconds

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3 years ago
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3 years ago
A car accelerates in the +x direction from rest with a constant acceleration of a1 = 1.76 m/s2 for t1 = 20 s. At that point the
alex41 [277]

Answer:

(a)v_1 = a_1t_1 = 1.76 t_1

(b) It won't hit

(c) 110 m

Explanation:

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(b) The velocity of the car before the driver begins braking is

v_1 = 1.76*20 = 35.2m/s

The driver brakes hard and come to rest for t2 = 5s. This means the deceleration of the driver during braking process is

a_2 = \frac{\Delta v_2}{\Delta t_2} = \frac{v_2 - v_1}{t_2} = \frac{0 - 35.2}{5} = -7.04 m/s^2

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Also the distance from start to where the driver starts braking is

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8 0
3 years ago
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3 years ago
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Make m the subject of equation 1

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Constant: 9.8 m/s²

Substitute these value into equation 2

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Where W' = weight of the alien on planet 9, g' = acceleration due to gravity on planet 9.

make g' the subject of the equation

g' = W'/m............ Equation 4

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Substitute into equation 4

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g' = 2.45 m/s²

5 0
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