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Elden [556K]
3 years ago
10

A uniform rod is set up so that it can rotate about an axis at perpendicular to one of its ends. The length and mass of the rod

are 0.899 m0.899 m and 1.13 kg1.13 kg , respectively. A force of constant magnitude FF acts on the rod at the end opposite the rotation axis. The direction of the force is perpendicular to both the rod's length and the rotation axis. Calculate the value of FF that will accelerate the rod from rest to an angular speed of 6.61 rad/s6.61 rad/s in 8.97 s8.97 s .
Physics
1 answer:
allsm [11]3 years ago
4 0

Answer:

Explanation:

angular acceleration α = 6.61 / 8.97 = .737 rad / s²

moment of inertia = 1/3  m L²

= 1/3  x 1.13 x .899²

= .3045  kg m²

torque applied = moment of inertia x angular acceleration

= .737 x .3045 = .2244

If be the force

Torque

= F x .899 = .2244

F = .25 N ,

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If a hot air balloon is seen rising above the top of a hill, what is the top of the hill called? Question 2 options: delta point
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If a hot air balloon has shown rising above the top of a hill, the top of the hill is considered as a reference point.

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A hungry rat is placed in a maze. It walks the following path to find a piece of cheese. 4.0m N, 7.5 m E, 6.8 m S, 3.7 m E, 3.6
storchak [24]

Answer:

Explanation:

We shall take the help of vector form of displacement . Taking east as i and north as j

4.0m N = 4 j

7.5 m E = 7.5 i

6.8 m S = - 6.8 j

3.7 m E, = 3.7 i

3.6 m S = - 3.6 j

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5.6 m W = - 5.6 i

4.4 m S = - 4.4 j

4.9 m W = -  4.9 i

Total displacement = 4j +7.5 i -6.8j+3.7i-3.6j-5.3i+3.7j-5.6i-4.4j-4.9i

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Tanθ = 7.1/ 4.6

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= 49.5 m

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