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Grace [21]
3 years ago
14

A constant torque of 26.6 N · m is applied to a grindstone for which the moment of inertia is 0.162 kg · m2 . Find the angular s

peed after the grindstone has made 11.8 rev. Assume the grindstone starts from rest.
Physics
2 answers:
ozzi3 years ago
7 0

Answer:

156.1 rad/s = 24.8 rev/s

Explanation:

Torque = Momentum of inertial × radial acceleration = Iα

τ = 26.6 N.m

I = 0.162 kg.m²

26.6 = 0.162 × α

α = 164.2 rad/s²

Using equations of motion,

θ = 11.8 rev = 11.8 × 2π = 74.2 rad

w₀ = 0 rad/s (since the grindstone starts from rest)

w = ?

α = 164.2 rad/s²

w² = w₀² + 2(α)(θ)

w² = 0² + (2×164.2)(74.2)

w = 156.1 rad/s = 24.8 rev/s

Hope this Helps!!!

Setler79 [48]3 years ago
6 0

Answer:

Explanation:

wi = 0 rad/s (starts from rest)

Torque, T= 26.6 N.m

Moment of inertia, I = 0.162 kg.m2

Theta = 11.8 rev

Converting rev to rad,

1 rev = 2pi rad

11.8 rev × 2pi rad/1 rev

= 74.14 rad

Using the formula,

T = I × a

Where,

a = angular acceleration

a = 26.6/0.162

= 164.2 rad/s^2

Using equations of circular rmotion,

wf^2 = wi^2 + 2a × theta

= 0 + 2 × 164.2 × 74.14

= 12174.05

wf = sqrt(12174.05)

= 110.34 rad/s.

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What is the combined resistance for two resistors with resistances of 10 Ohms and 23.7 Ohms if they are connected in series
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Explanation:

Since there are two resistors connected in series, the total resistance (Rtotal) of the circuit is the sum of each resistance.

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Hence, Rtotal = 10Ω + 23.7Ω

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Unpolarized light passes through two polarizers whose transmission axes are at an angle # with respect to each other. What shoul
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Answer:

63.4^{\circ}

Explanation:

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When the (now polarized) light hits the second polarizer, whose axis of polarization is rotated by an angle \theta with respect to the first one, the intensity of the light coming out is

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3 years ago
Firemen are shooting a stream of water at a burning building. A high-pressure hose shoots out the water with a speed of 26.0 m/s
alekssr [168]

Answer:

a) θ = 58.3º

b) vfh = 13.7 m/s

c) g = -9.8 m/s2

d) h = 22.2 m

e) vfb = 15.5 m/s

Explanation:

a)

  • Assuming that gravity is the only influence that causes an acceleration to the water, due to it is always downward, since both directions are independent each other, in the horizontal direction, the water moves at a constant speed.
  • Since the velocity vector has a magnitude of 26.0 m/s, we can find its horizontal component as follows:
  • vₓ₀ = v * cos θ (1)
  • where θ is the angle between the water and the horizontal axis (which we define as the x-axis, being positive to the right).
  • Applying the definition of average velocity, taking the end of the hose like the origin, and making t₀ = 0, we can write the following expression:

        x_{f} = v_{ox} * t = v_{o} * cos \theta * t  (2)

  • Replacing by the givens of xf = 41.0m, t = 3.00 s, and v=26.0 m/s, we can solve for the angle of elevation θ, as follows:

        cos \theta = \frac{x_{f} }{v*t} = \frac{41.0m}{26.0m/s*3.00s} = 0.526 (3)

  • ⇒θ = cos⁻¹ (0.526) = 58.3º (4)

b)

  • At the highest point in its trajectory, just before starting to fall, the vertical component of the velocity is just zero.
  • Since the horizontal component keeps constant during all the journey, we can conclude that the speed at this point is just v₀ₓ, that we can find easily from (1) replacing by the values of v and cos θ, as follows:
  • vₓ₀ = v * cos θ = 26.0 m/s * 0.526 = 13.7 m/s. (5)

c)

  • At any point in the trajectory, the only acceleration present is due to the action of gravity, which accepted value is -9.8 m/s2 (taking the upward direction on the vertical y-axis as positive)

d)

  • Since we know the time when the water strikes the building, it will be the same for the vertical movement, so, we can use the kinematic equation for vertical displacement, as follows:

       \Delta y = v_{oy} * t - \frac{1}{2} *g*t^{2} (6)

  • Our only unknown remains v₀y, which can be obtained in the same way than the horizontal component:
  • v₀y = v * sin θ = 26.0 m/s * 0.85 = 22.1 m/s (7)
  • Replacing (7) in (6), we get:

       \Delta y = 22.1 m/s* 3.0s - \frac{1}{2} *9.8m/s2*(3.00s)^{2} = 22.2 m (8)

e)

  • When the water hits the building the velocity vector, has two components, the horizontal vₓ and the vertical vy.
  • The horizontal component, since it keeps constant, is just v₀x:
  • v₀ₓ = 13.7 m/s
  • The vertical component can be found applying the definition of acceleration (g in this case), solving for the final velocity, as follows:

       v_{fy} = v_{oy} - g*t  (9)

  • Replacing by the time t (a given), g, and  v₀y from (7), we can solve (9) as follows:

       v_{fy} = 22.1 m/s - 9.8m/s2*3.00s = -7.3 m/s  (10)

  • Since we know the values of both components (perpendicular each other), we can find the magnitude of the velocity vector (the speed, i.e. how fast is it moving), applying the Pythagorean Theorem to v₀ₓ and v₀y, as follows:

       v_{f} = \sqrt{(13.7m/s)^{2} +(-7.3m/s)^{2}} = 15.5 m/s (11)

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