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Afina-wow [57]
2 years ago
6

The figure above shows a rod that is fixed to a horizontal surface at pivot P The rod is initially rotating

Physics
1 answer:
Lisa [10]2 years ago
5 0

The angular speed is decreasing and direction of rotation clockwise of the rod immediately after time t.

<h3></h3><h3>What is angular speed ?</h3>

The rate of change of angular displacement is defined as angular speed.  It is stated as follows:

ω = θ t

Where,

θ is the angle of rotation,

t is the time

ω is the angular velocity

The torque is found as;l

\rm  T = F \times r

If the force is acting on the rod  from the three point is the same, the value of the torque is depends upon the radius or the perpendicular distance.

The perpendicular distance of the right force is grater. Hence, the force acting on the right side is more, and the rod will rotate clockwise.

Both the forces are acting downwards. Thus, the resultant force is the less due to which the speed is increasing.

Hence, the angular speed is decreasing and direction of rotation clockwise of the rod immediately after time t.

To learn more about the angular speed, refer to the link;

brainly.com/question/9684874

#SPJ1

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The acceleration due to gravity on the Moon's surface is
Molodets [167]

Answer:

50 lb

Explanation:

Given,

The weight of astronaut's life support backpack on Earth (w) = 300 lb

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Acceleration due to gravity on Moon = g'

g'=\frac{g}{6}

We know that weight of an object on Earth is,

w = m\times g

m = \frac{w}{g}

Similarly, weight on Moon will be

w' = m\times g'

w' = \frac{w}{g}\times\frac{g}{6}

w' = \frac{300}{6}

w' = 50

Thus the astronaut's life support backpack will weigh 50 lb on Moon.

7 0
3 years ago
Through what potential difference would you need to accelerate an alpha particle, starting from rest, so that it will just reach
Svetlanka [38]

Answer:

\Delta V    = 1.8 \times 10^7 V

Explanation:

GIVEN

diameter = 15 fm  =15 \times 10^{-15}m

we use here energy conservation

K_{i}+U_{i} =K_{f}+U_{f}

there will be some initial kinetic  energy but after collision kinetic energy will zero

K_{i} + 0 = 0 + \frac{1}{4 \pi \epsilon _{0}} \frac{(2e)(92e)}{7.5 \times 10^{-15}}

on solving these equations we get kinetic energy initial

KE_{i} = 5.65\times 10 ^{-12} \times \frac  {1 eV}{1.6 \times 10^{-19}}

KE_{i} = 35.33 J ..............(i)

That is, the alpha particle must be fired with 35.33 MeV of kinetic energy. An alpha particle with charge q = 2  e

and gains kinetic energy K  =e∆V  ..........(ii)

 by accelerating through a potential difference ∆V

Thus the alpha particle will

just reach the {238}_U nucleus after being accelerated through a potential difference  ∆V

equating (i) and second equation we get

e∆V  = 35.33 Me V

\Delta V = \frac{35.33}{2}  MV\\\Delta V    = 1.8 \times 10^7 V

7 0
3 years ago
A11) A solenoid of length 18 cm consists of closely spaced coils of wire wrapped tightly around a wooden core. The magnetic fiel
vitfil [10]

Answer:

A

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B= frac{\mu_0}{2R}

As I is constant

\frac{B2}{B1} = \frac{R1}{R2}

B2 = 2mT*\frac{18}{21}

B2 = 1.714mT

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3 years ago
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