Answer:
1.37 rad/s
Explanation:
Given:
Total length of the tape is,
m
Total time of run is,
hours
We know, 1 hour = 3600 s
So, 2.1 hours = 2.1 × 3600 = 7560 s
So, total time of run is,
s
Inner radius is, ![r = 10\ mm = 0.01\ m ](https://tex.z-dn.net/?f=r%20%3D%2010%5C%20mm%20%3D%200.01%5C%20m%0A)
Outer radius is, ![R = 47\ mm = 0.047\ m ](https://tex.z-dn.net/?f=R%20%3D%2047%5C%20mm%20%3D%200.047%5C%20m%0A)
Now, linear speed of the tape is, ![v =\frac{d}{t}=\frac{297}{7560}=0.039\ m/s ](https://tex.z-dn.net/?f=v%20%3D%5Cfrac%7Bd%7D%7Bt%7D%3D%5Cfrac%7B297%7D%7B7560%7D%3D0.039%5C%20m%2Fs%0A)
Let the same angular speed be
.
Now, average radius of the reel is given as the sum of the two radii divided by 2.
So, average radius is, ![R_{avg}=\frac{R+r}{2}=\frac{0.047+0.01}{2}=\frac{0.057}{2}=0.0285\ m ](https://tex.z-dn.net/?f=R_%7Bavg%7D%3D%5Cfrac%7BR%2Br%7D%7B2%7D%3D%5Cfrac%7B0.047%2B0.01%7D%7B2%7D%3D%5Cfrac%7B0.057%7D%7B2%7D%3D0.0285%5C%20m%0A)
Now, common angular speed is given as the ratio of linear speed and average radius of the tape. So,
![\omega=\dfrac{v}{R_{avg}}\\\\\\\omega=\dfrac{0.039}{0.0285}\\\\\\\omega=1.37\ rad/s ](https://tex.z-dn.net/?f=%5Comega%3D%5Cdfrac%7Bv%7D%7BR_%7Bavg%7D%7D%5C%5C%5C%5C%5C%5C%5Comega%3D%5Cdfrac%7B0.039%7D%7B0.0285%7D%5C%5C%5C%5C%5C%5C%5Comega%3D1.37%5C%20rad%2Fs%0A)
Therefore, the common angular speed of the reels is 1.37 rad/s.
Answer:
a. 32.67 rad/s² b. 29.4 m/s²
Explanation:
a. The initial angular acceleration of the rod
Since torque τ = Iα = WL (since the weight of the rod W is the only force acting on the rod , so it gives it a torque, τ at distance L from the pivot )where I = rotational inertia of uniform rod about pivot = mL²/3 (moment of inertia about an axis through one end of the rod), α = initial angular acceleration, W = weight of rod = mg where m = mass of rod = 1.8 kg and g = acceleration due to gravity = 9.8 m/s² and L = length of rod = 90 cm = 0.9 m.
So, Iα = WL
mL²α/3 = mgL
dividing through by mL, we have
Lα/3 = g
multiplying both sides by 3, we have
Lα = 3g
dividing both sides by L, we have
α = 3g/L
Substituting the values of the variables, we have
α = 3g/L
= 3 × 9.8 m/s²/0.9 m
= 29.4/0.9 rad/s²
= 32.67 rad/s²
b. The initial linear acceleration of the right end of the rod?
The linear acceleration at the initial point is tangential, so a = Lα = 0.9 m × 32.67 rad/s² = 29.4 m/s²
Answer:
131.25
Explanation:
i worked it out on a diffrent sheet so its hard to explain
Answer:
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Explanation:
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Answer:
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