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garri49 [273]
3 years ago
9

An unruly student with a spitwad (a lump of wet paper) of mass 20 g in his pocket finds himself in the school library where ther

e is a ceiling fan overhead. He relieves his boredom by throwing the spitwad up at the ceiling fan where it collides with, and sticks to, the end of one of the blades of the stationary ceiling fan. Its horizontal velocity vector is perpendicular to the long axis of the blade. If the fan is free to rotate (no friction at all) and has moment of inertia I=1.4kgm2 , if the spitwad has horizontal velocity 4 m/s, and if the spitwad sticks to the fan blade at a distance of 0.6 m from the rotation axis of the fan, how much time will it take the fan to move through one complete revolution after the spitwad hits it (closest answer)?
a. 1min
b. 2min
c. 3min
d. 4min
e. 5min
f. 6min
Physics
1 answer:
jeka943 years ago
6 0

Answer:

T = 188.5 s, correct is  C

Explanation:

This problem must be worked on using conservation of angular momentum. We define the system as formed by the fan and the paper, as the system is isolated, the moment is conserved

         

initial instant. Before the crash

        L₀ = r m v₀ + I₀ w₀

the angular speed of the fan is zero w₀ = 0

final instant. After the crash

        L_f = I₀ w + m r v

        L₀ = L_f

        m r v₀ = I₀ w + m r v

angular and linear velocity are related

        v = r w

        w = v / r

        m r v₀ = I₀ v / r + m r v

         m r v₀ = (I₀ / r + mr) v

       v = \frac{m}{\frac{I_o}{r}  +mr} \ r v_o

let's calculate

       v = \frac{0.020}{\frac{1.4}{0.6  } + 0.020 \ 0.6  } \ 0.6 \ 4

       v = \frac{0.020}{2.345} \ 2.4

       v = 0.02 m / s

         

To calculate the time of a complete revolution we can use the kinematics relations of uniform motion

        v = x / T

         T = x / v

the distance of a circle with radius r = 0.6 m

         x = 2π r

we substitute

         T = 2π r / v

let's calculate

         T = 2π 0.6/0.02

         T = 188.5 s

reduce

         t = 188.5 s ( 1 min/60 s) = 3.13 min

correct is  C

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Rudik [331]

Answer with Explanation:

We are given that

F_1=6.9 N

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We have to find the direction and magnitude of the third force acting on the object.

Resultant force,F=\sqrt{F^2_1+F^2_2}

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The object moves with constant velocity .Therefore, net force on object is equal to zero

So,Third force,F_3=F=8.24 N

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\theta=tan^{-1}(\frac{4.5}{6.9})=33.02^{\circ}

Angle lies in second quadrant because the direction of third force is opposite to the direction of the resultant force of F1 and F2.

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Answer:

a) 5.64 C

b) 3.5*10¹⁹ protons

Explanation:

a)

  • Since the four metallic objects are identical, and total charge must be conserved, this means that after brought simultaneously into contact so that each touches the others, once separated, total charge must be the same than before being brought in contact.
  • But due they are identical, after charges were able to transfer freely between them, the four objects must have the same final charge, i.e. the fourth part of the total charge, as follows:

       Q_{n} = \frac{Q_{tot}}{4} = \frac{22.57C}{4} = 5.64 C  (1)

b)

  • This charge will be divided between n protons, since the charge is positive.
  • Since each proton carries a charge equal to the elementary charge e, which value is 1.6*10⁻¹⁹ C, we can find the number of protons in excess, doing the following calculation:
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