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cluponka [151]
3 years ago
8

When you drop an object from a certain height, it takes time T to reach the ground with no air resistance. If you dropped it fro

m three times that height, how long (in terms of T) would it take to reach the ground?
Physics
1 answer:
alexandr402 [8]3 years ago
5 0

Answer:

Explanation:

Given

When we drop an object from height , suppose h

it takes time T

using equation of motion

h=ut+\frac{1}{2}at^2

where  

h=displacement

u=initial\ velocity

a=acceleration

t=time

here u=0 because it dropped from a certain height

h=\frac{1}{2}gT^2

T=\sqrt{\frac{2h}{g}}

When height is increases to three times of original height

i.e. h'=3 h

then time period becomes

T'=\sqrt{\frac{2\times 3h}{g}}

T'=\sqrt{3}T

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In freely falling body, there is no force acting on it other than the force of gravity (g).
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3 years ago
Consider a machine of mass 70 kg mounted to ground through an isolation system of total stiffness 30,000 N>m, with a measured
kvv77 [185]

Answer:

a)0.0229 m

b)0.393 rad

c)1.57

d)707.6 N

e)0.298 m/s

Explanation:

Given:

  • Mass of the machine, m=70 kg
  • Stiffness of the system, k=30000 N/m
  • Damping ratio=0.2
  • Damping force, F=450 N
  • Angular velocity \omega=13\ \rm rad/s

a)We know that the amplitude X at steady state is given by

X=\dfrac{\dfrac{F_0}{m}}{\sqrt{\omega_n^2-\omega^2)^2 +(2\rho \omega_n\omega)^2}}\\

Where

  • \omega_n=\sqrt{\dfrac{k}{m}}\\\\=\sqrt{\dfrac{30000}{70}}\\\\=20.7\ \rm rad/s
  • \omega=13\ \rm rad/s
  • \rho=0.2
  • F_0=450\ \rm N
  • m=70\ \rm kg

X=\dfrac{\dfrac{450}{70}}{\sqrt{20.7^2-13^2)^2 +(2\times 0.2\times20.7\times13)^2}}\\[tex]X=0.0229\ \rm m

b) The phase shift of the motion is given by

\tan\phi=\dfrac{2\rho \omega_n \omega }{\omega_n^2-\omega^2}\\\\\dfrac{2\times0.2\times20.7\times13 }{20.7^2-\13^2}\\\\\phai=0.393\\

c)Transmissibility ratio is given by

T.R.=\sqrt{\dfrac{1+(2\rho r)^2}{(1-r^2)^2+{(2\rho r)^2}}}\\\\T.R.=\sqrt{\dfrac{1+(2\times0.2\times0.628)^2}{(1-0.628^2)^2+{(2\times0.2\times0/628)^2}}}\\\\=1.57

d)The magnitude of the force transmitted to the ground is

F_T=(T.R)\times F_0\\\\=450\times1.57\\\\=707.6\ \rm N

e)The maximum velocity is given by V_{max}

V_{max}=\omega A_0\\\\=13\times 0.0229\\\\=0.298\ \rm m/s

6 0
3 years ago
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8 0
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An empty 230 kg elevator accelerates upward
elena-s [515]

Answer:

7.2 as used in the equation

3 0
3 years ago
A car tire rotates at a constant angular velocity of 3.5 rotations during a time interval of 0.75 seconds. What is the angular s
Andru [333]

Answer : \theta = 2.625\ radians

Explanation :  

Given that,

Angular velocity \omega = 3.5\ rad/s

Time t = 0.75 s

We know , the formula of angular velocity

\omega = \dfrac{\theta}{t}

\theta = \omega \times t

\theta = 3.5\ rad/s\times0.75\ s

The angular speed of the tire is

\theta = 2.625\ radians

Hence, this is the required solution.



4 0
3 years ago
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