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IRISSAK [1]
3 years ago
14

A rope is used to pull a 4.36 kg block at constant speed 4.33 m along a horizontal floor. The force on the block from the rope i

s 7.97 N and directed 26.2° above the horizontal. What are (a) the work done by the rope's force, (b) the increase in thermal energy of the block-floor system, and (c) the coefficient of kinetic friction between the block and floor?
Physics
1 answer:
geniusboy [140]3 years ago
4 0

Answer:

Explanation:

The horizontal component of force applied = 7.97 cos 26.2 = 7.15 N.

Vertical component in upward direction = 7.97 sin 26.2 = 3.52 N.

Since the body is moving with uniform velocity, friction force will equalize the external horizontal component = 7.15 N.

So frictional force  = 7.15 N.

a) Work done by rope force per second = force in horizontal direction x displacement per second = 7.15 x 4.33 = 30.95 J

b) Increase in thermal energy per second will be due to negative work done by frictional force = work done by external force = 30.95 J.

c) Normal force acting downwards = weight - vertical component of external force  = 4.36 x 9.8 - 3.52 = 39.21 N

coefficient of friction = friction force / normal force = 7.15 / 39.21 = 0.18

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Kaylis [27]

The magnitudes of the forces that the ropes must exert on the knot connecting are :

  • F₁ = 118 N
  • F₂ = 89.21 N
  • F₃ = 57.28 N

<u>Given data :</u>

Mass ( M ) = 12 kg

∅₂ = 63°

∅₃ = 45°

<h3>Determine the magnitudes of the forces exerted by the ropes on the connecting knot</h3><h3 />

a) Force exerted by the first rope = weight of rope

∴  F₁ = mg

     = 12 * 9.81 ≈  118 kg

<u>b) Force exerted by the second rope </u>

applying equilibrium condition of force in the vertical direction

F₂ sin∅₂ + F₃ sin∅₃ - mg = 0  ---- ( 1 )

where: F₃ = ( F₂ cos∅₂ / cos∅₃ ) --- ( 2 )  applying equilibrium condition of force in the horizontal direction

Back to equation ( 1 )

F₂ =  [ ( mg / cos∅₂ ) / tan∅₂ + tan∅₃ ]

   = [ ( 118 / cos 63° ) / ( tan 63° + tan 45° ) ]

   = 89.21 N

<u />

<u>C ) </u><u>Force </u><u>exerted by the</u><u> third rope </u>

Applying equation ( 2 )

F₃ = ( F₂ cos∅₂ / cos∅₃ )

    = ( 89.21 * cos 63 / cos 45 )

    = 57.28 N

Hence we can conclude that The magnitudes of the forces that the ropes must exert on the knot connecting are :

F₁ = 118 N, F₂ = 89.21 N, F₃ = 57.28 N

Learn more about  static equilibrium : brainly.com/question/2952156

6 0
2 years ago
A proton moves perpendicular to a uniform magnetic field B at a speed of 2.30 107 m/s and experiences an acceleration of 2.50 10
Katen [24]

Answer:

B = 6.18 10⁻⁶ T  

the magnetic field is in the negative direction of the y axis

Explanation:

The magnetic force is given by

         F = q v x B

as in the exercise indicate that the velocities perpendicular to the magnetic field,

         F = q v B

Newton's second law is

         F = m a

let's substitute

         q v B = m a

         B = m a / q v

let's calculate

         B = 9.1 10⁻³¹ 2.50 10¹³ / (1.6 10⁻¹⁹ 2.30 10⁷)

         B = 6.18 10⁻⁶ T

The direction of the field can be obtained with the right hand rule, where the thumb points in the direction of the velocity, the fingers extended in the direction of the magnetic field and the palm in the direction of the force for a positive charge.

In the exercise indicate that the velocity is the z axis

the acceleration and therefore the force in the x axis

therefore the magnetic field is in the negative direction of the y axis

7 0
3 years ago
Samuel adds a teaspoon of salt to a glass of water. He notices that the salt disappears. Samuel takes a dip to discover that the
Tpy6a [65]
It is a chemical change because the salt is dissolving.
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A particular circuit toolbox contains only 80 Ω resistors and switches, which can be open or closed. Construct a circuit, fillin
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Answer:

Here we need to make parallel connection of two 80 ohm resistors to achieve 40 ohm net resistance.

Explanation:

As we know that the resistances in series add up directly and here we are given with only the resistors of 80 Ω.

So when we connect two resistors of 80 ohm in parallel we get the resultant of 40 ohm.

Mathematically:

\frac{1}{R_p} =\frac{1}{R} +\frac{1}{R}

\frac{1}{40} =\frac{1}{R} +\frac{1}{R}

\frac{1}{40} =\frac{2}{R}

R=80\Omega gives us the only combination of two resistors in parallel.

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A ball is thrown straight upward. When it has reached the highest point in its motion, and is momentarily stopped, its accelerat
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