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gavmur [86]
3 years ago
13

Consider two metallic rods mounted on insulated supports. One is neutral, another positively charged. The charged rod is brought

close to the neutral one (no touching!) and at this time the neutral rod was briefly grounded (for instance, touched by your hand).
Part (a) What would be resulting charge (if any) on initially neutral rod?

1) There is not enough information given to be able to answer.
2) It will be positively charged.
3) It will be negatively charged.
4) It will remains neutral.
Physics
1 answer:
serious [3.7K]3 years ago
8 0

Answer:

The resulting charge on the initially neutral rod will be  3.) it will be negatively charged.

Explanation:

If the neutral rod is brought close to the neutral one and if the neutral rod is touching the ground then the positive charges will attract electrons from the ground on to the rod.

The Earth (or ground) has electrons in abundance and electrons can be taken from of fed into the Earth without having any significant impact on the Earth's electric field.

So. the resulting charge on the initially neutral rod will be  3.) it will be negatively charged.

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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

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3 years ago
Describe the political actions that led to successful conservation in both stories.
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Answer:

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Explanation:

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A 5kg box rests on a table. a 3 kg box rests on top of the 5 kg box. what is the normal force from the table?
Murrr4er [49]

==>  The total mass resting on the table is (5 kg + 3 kg) = 8 kg.

==>  The total weight of that mass is (8 kg) x (9.8 m/s) = 78.4 newtons

==>  The boxes are stacked.  So the table doesn't know if the weight on it is coming from one box, 2 boxes, 3 boxes, or 100 boxes in a stack.  The table only knows that there is a downward force of 78.4 newtons on it.

==>  The table stands in a Physics classroom, and it soaks up everything it hears there.  It knows that every action produces an equal and opposite reaction, and that forces always occur in pairs.  

Ever since the day it was only a pile of lumber out behind the hardware store in the rain, the table has known that in order to maintain the good  reputation of tables all over the world, it must resist the weight of anything placed upon it with an identical upward force.  This is the normal thing for all good tables to do, up to the ultimate structural limit of their materials and construction, and it is known as the "normal force".

So the table in your question provides a normal force of 78.4 newtons. (d)

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Answer:

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Explanation:

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The correct answer is 4
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