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Fantom [35]
2 years ago
11

A 2.8 kg grinding wheel is in the form of a solid cylinder of radius 0.1 m. a) What constant torque will bring it from rest to a

n angular velocity of 120 rad/s in 2.5 s. b) Through what angle has it turned during that time? c) What is the work done by the torque?
Physics
1 answer:
kogti [31]2 years ago
7 0

Answer:

a) τ =  0.672 N m , b) θ = 150 rad , c) W = 100.8 J

Explanation:

a) for this part let's start by finding angular acceleration, when the angular velocity stops it is zero (w = 0)

       w = w₀ + α t

       α = -w₀ / t

       α = 120 / 2.5

       α = 48 rad / s²

The moment of inertia of a cylinder is

       I = ½ M R²

Let's calculate the torque

      τ = I α

      τ = ½ M R² α

      τ = ½ 2.8 0.1² 48

      τ =  0.672 N m

b) we look for the angle by kinematics

      θ = w₀ t + ½ α t2

      θ = ½ α t²

      θ = ½ 48 2.5²

      θ = 150 rad

c) work in angular movement

      W = τ θ

      W = 0.672 150

      W = 100.8 J

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\theta =  \frac{S}{r}
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\theta =  \frac{2600 m}{0.35 m} =7428 rad

2) If we put larger tires, with radius r=0.60 m, the angular displacement will be smaller. We can see this by using the same formula. In fact, this time we have:
\theta =  \frac{2600 m}{0.60 m}=4333 rad
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3 years ago
A 2.00 kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0300 m . The spring has
iogann1982 [59]

Answer:

v = 0.41 m/s

Explanation:

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  • At any point, the total mechanical energy is the sum of the kinetic energy plus the elastic potential energy.
  • So, we can write the following general equation, taking the initial and final values of the energies:

       \Delta K + \Delta U = W_{ffr}  (1)

  • Since the block and spring start at rest, the change in the kinetic energy is just the final kinetic energy value, Kf.
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  • The change in the potential energy, can be written as follows:

       \Delta U = U_{f}  - U_{o}  = \frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (3)

       where k = force constant = 815 N/m

       xf = final displacement of the block = 0.01 m (taking as x=0 the position

      for the spring at equilibrium)

      x₀ = initial displacement of  the block = 0.03 m

  • Regarding the work done by the force of friction, it can be written as follows:

       W_{ffr} = - \mu_{k}* F_{n} * \Delta x  (4)

       where μk = coefficient of kinettic friction, Fn = normal force, and Δx =

       horizontal displacement.

  • Since the surface is horizontal, and no acceleration is present in the vertical direction, the normal force must be equal and opposite to the force due to gravity, Fg:
  • Fn = Fg= m*g (5)
  • Replacing (5) in (4), and (3) and (4) in (1), and rearranging, we get:

        \frac{1}{2} * m* v^{2} = W_{ffr} - \Delta U = W_{ffr} - (U_{f} -U_{o})  (6)

        \frac{1}{2} * m* v^{2} = (- \mu_{k}* m*g* \Delta x)  -\frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (7)

  • Replacing by the values of m, k, g, xf and x₀, in (7) and solving for v, we finally get:

    \frac{1}{2} * 2.00 kg* v^{2}  = (-0.4*2.00 kg*9.8m/s2*0.02m) +( (\frac{1}{2} *815 N/m)* (0.03m)^{2} - (0.01m)^{2}) = -0.1568 J + 0.326 J (8)

  • v =\sqrt{(0.326-0.1568}  =  0.41 m/s  (9)
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evablogger [386]

Answer:

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A milliammeter has an internal resistance of 5ohms and
Vikentia [17]

Answer:

Rs = 0.02008 Ω = 20.08 mΩ

Explanation:

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Rs = \frac{I_gR_g}{I - I_g}

where,

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I_g = current range of ammeter = 20 mA = 0.02 A

I = Required range of ammeter = 5 A

R_g = Resistance of ammeter = 5 ohms

Therefore,

R_s = \frac{(0.02\ A)(5\ ohms)}{5\ A-0.02\ A}

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