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alukav5142 [94]
3 years ago
11

A disk 7.90 cm in radius rotates at a constant rate of 1 190 rev/min about its central axis. (a) Determine its angular speed. 12

4.58 Correct: Your answer is correct. rad/s (b) Determine the tangential speed at a point 2.98 cm from its center. 3.71 Correct: Your answer is correct. m/s (c) Determine the radial acceleration of a point on the rim. magnitude 1.23 Correct: Your answer is correct. km/s2 direction toward the center Correct: Your answer is correct. (d) Determine the total distance a point on the rim moves in 2.06 s. 20.28 Correct: Your answer is correct. m
Physics
1 answer:
Tanya [424]3 years ago
8 0

Answer:

124.62\ \text{rad/s}

3.71\ \text{m/s}

1.23\ \text{km/s}^2

20.28\ \text{m}

Explanation:

r = Radius of disk = 7.9 cm

N = Number of revolution per minute = 1190 rev/minute

Angular speed is given by

\omega=N\dfrac{2\pi}{60}\\\Rightarrow \omega=1190\times \dfrac{2\pi}{60}\\\Rightarrow \omega=124.62\ \text{rad/s}

The angular speed is 124.62\ \text{rad/s}

r = 2.98 cm

Tangential speed is given by

v=r\omega\\\Rightarrow v=2.98\times 10^{-2}\times 124.62\\\Rightarrow v=3.71\ \text{m/s}

Tangential speed at the required point is 3.71\ \text{m/s}

Radial acceleration is given by

a=\omega^2r\\\Rightarrow a=124.62^2\times 7.9\times 10^{-2}\\\Rightarrow a=1226.88\approx 1.23\ \text{km/s}^2

The radial acceleration is 1.23\ \text{km/s}^2.

t = Time = 2.06 s

Distance traveled is given by

d=vt\\\Rightarrow d=\omega rt\\\Rightarrow d=124.62\times 7.9\times 10^{-2}\times 2.06\\\Rightarrow d=20.28\ \text{m}

The total distance a point on the rim moves in the required time is 20.28\ \text{m}.

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Detailed solution is given below

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Does every light source emit only one type of light?
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Yes it is possible. Spectrum of emitted light depends upon the chemical composition of the source. and the way of its excitation. a clear example to us is that of sun.

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A puck of mass 0.110 kg slides across ice in the positive x-direction with a kinetic friction coefficient between the ice and pu
lara [203]

Answer:

a) Ffr = -0.18 N

b) a= -1.64 m/s2

c) t = 9.2 s

d) x = 68.7 m.

e) W= -12.4 J

f) Pavg = -1.35 W

g) Pinst = -0.72 W

Explanation:

a)

  • While the puck slides across ice, the only force acting in the horizontal direction, is the force of kinetic friction.
  • This force is the horizontal component of the contact force, and opposes to the relative movement between the puck and the ice surface, causing it to slow down until it finally comes to a complete stop.
  • So, this force can be written as follows, indicating with the (-) that opposes to the movement of the object.

       F_{frk} = -\mu_{k} * F_{n} (1)

       where μk is the kinetic friction coefficient, and Fn is the normal force.

  • Since the puck is not accelerated in the vertical direction, and there are only two forces acting on it vertically (the normal force Fn, upward, and  the weight Fg, downward), we conclude that both must be equal and opposite each other:

      F_{n} = F_{g} = m*g (2)

  • We can replace (2) in (1), and substituting μk by its value, to find the value of the kinetic friction force, as follows:

       F_{frk} = -\mu_{k} * F_{n} = -0.167*9.8m/s2*0.11kg = -0.18 N (3)

b)

  • According Newton's 2nd Law, the net force acting on the object is equal to its mass times the acceleration.
  • In this case, this net force is the friction force which we have already found in a).
  • Since mass is an scalar, the acceleration must have the same direction as the force, i.e., points to the left.
  • We can write the expression for a as follows:

        a= \frac{F_{frk}}{m} = \frac{-0.18N}{0.11kg} = -1.64 m/s2  (4)

c)

  • Applying the definition of acceleration, choosing t₀ =0, and that the puck comes to rest, so vf=0, we can write the following equation:

        a = \frac{-v_{o} }{t} (5)

  • Replacing by the values of v₀ = 15 m/s, and a = -1.64 m/s2, we can solve for t, as follows:

       t =\frac{-15m/s}{-1.64m/s2} = 9.2 s (6)

d)

  • From (1), (2), and (3) we can conclude that the friction force is constant, which it means that the acceleration is constant too.
  • So, we can use the following kinematic equation in order to find the displacement before coming to rest:

        v_{f} ^{2} - v_{o} ^{2} = 2*a*\Delta x  (7)

  • Since the puck comes to a stop, vf =0.
  • Replacing in (7) the values of v₀ = 15 m/s, and a= -1.64 m/s2, we can solve for the displacement Δx, as follows:

       \Delta x  = \frac{-v_{o}^{2}}{2*a} =\frac{-(15.0m/s)^{2}}{2*(-1.64m/s2} = 68.7 m  (8)

e)

  • The total work done by the friction force on the object , can be obtained in several ways.
  • One of them is just applying the work-energy theorem, that says that the net work done on the object is equal to the change in the kinetic energy of the same object.
  • Since the final kinetic energy is zero (the object stops), the total work done by friction (which is the only force that does work, because the weight and the normal force are perpendicular to the displacement) can be written as follows:

W_{frk} = \Delta K = K_{f} -K_{o} = 0 -\frac{1}{2}*m*v_{o}^{2} =-0.5*0.11*(15.0m/s)^{2}   = -12.4 J  (9)

f)

  • By definition, the average power is the rate of change of the energy delivered to an object (in J) with respect to time.
  • P_{Avg} = \frac{\Delta E}{\Delta t}  (10)
  • If we choose t₀=0, replacing (9) as ΔE, and (6) as Δt, and we can write the following equation:

       P_{Avg} = \frac{\Delta E}{\Delta t} = \frac{-12.4J}{9.2s} = -1.35 W (11)

g)

  • The instantaneous power can be deducted from (10) as W= F*Δx, so we can write P= F*(Δx/Δt) = F*v (dot product)
  • Since F is constant, the instantaneous power when v=4.0 m/s, can be written as follows:

       P_{inst} =- 0.18 N * 4.0m/s = -0.72 W (12)

7 0
3 years ago
A mass hanging from a spring oscillates with a period of 0.35 s. Suppose the mass and spring are swung in a horizontal circle, w
Annette [7]

Answer:

66 rpm

Explanation:

The period of oscillation is given by

T=2\pi \sqrt{\frac {m}{k}}

\frac {k}{m}=\frac {4\pi^{2}}{T^{2}} where  T is time period of oscillation which is given as 0.35 s, k s spring constant and m is the mass of the object attached to the spring.

Also, net force is given by

Net force=m\omega^{2} L

\omega=\sqrt{\frac {k\triangle L}{mL}} where \triangle L is the elongation, L is original length, \omega is the angular velocity

Substituting the equation of \frac {k}{m} into the above we obtain

\omega=\sqrt {\frac {4\pi^{2}\triangle L}{T^{2} L}}

\omega=\sqrt {4\pi^{2}\times 0.15L}{0.35^{2}\times L}}=6.952763\approx 6.95 rad/s

6.95\times\frac {60 s}{2\pi rad}\approx 66 rpm

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Please help me find the answer ​
Hoochie [10]

Answer:

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27) 500 N

Explanation:

26.a) In a two pulley system, the load is attached to one of the pulleys.  The other pulley is attached to a fixed surface, as well as one end of the rope.  The other end of the rope goes around moving pulley, then around the fixed pulley.

26.b) Mechanical advantage is the ratio between the forces:

MA = load force / effort force

Efficiency is the ratio between the work:

e = work done on load / work done by effort

Work is force times distance.

e = (F load × d load) / (F effort × d effort)

Rearranging:

e = (F load / F effort) × (d load / d effort)

e = MA × (d load / d effort)

In a two pulley system, the load moves half the distance of the effort.  So the efficiency is:

e = (4/3) × (1/2)

e = 2/3

e = 66.7%

27) In a three pulley system, the load moves a third of the distance of the effort.

e = (F load / F effort) × (d load / d effort)

0.40 = (600 N / F) × (1/3)

F = 500 N

8 0
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