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3 years ago

Determine the unbalanced force necessary to accelerate a 2.60 kg object at a rate of 14.0 m/s².

Physics

1 answer:

slega [8]3 years ago

8 0

Answer:

Explanation:

F = ma

F = 2.60(14.0)

F = 36.4 N

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A cyclist is moving at a speed of 15 m/s. If the combined mass of the bike and person is 100 kg,

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The rotational inertia of a thin rod about one end is 1/3 ML2. What is the rotational inertia of the same rod about a point loca

zlopas [31]

Answer:

The value is $I = 0.0932 ML ^2$

Explanation:

From the question we are told that

The rotational inertia about one end is $I_R = \frac{1}{3} ML^2$

The location of the axis of rotation considered is $d = 0.4 L$

Generally the mass of the portion of the rod from the axis of rotation considered to the end of the rod is $0.4 M$

Generally the length of the rod from the its beginning to the axis of rotation consider is

$k = 1 - 0.4 L = 0.6L$

Generally the mass of the portion of the rod from the its beginning to the axis of rotation consider is

$m = 1- 0.4 M = 0.6 M$

Generally the rotational inertia about the axis of rotation consider for the first portion of the rod is

$I_{R1} = \frac{1}{3} (0.6 M )(0.6L)^2$

$I_{R1} = \frac{1}{3} (0.6 M )L^2 0.6^2$

Generally the rotational inertia about the axis of rotation consider for the second portion of the rod is

$I_{R2} = \frac{1}{3} (0.6 M )(0.6L)^2$

=> $I_{R2} = \frac{1}{3} (0.4 M )(0.4L)^2$

=> $I_{R2} = \frac{1}{3} (0.4 M )L^2 0.4^2$

Generally by the principle of superposition that rotational inertia of the rod at the considered axis of rotation is

$I = \frac{1}{3} (0.6 M )L^2 0.6^2 + \frac{1}{3} (0.4 M )L^2 0.4^2$

=> $I = \frac{1}{3} ML ^2 [0.6 * (0.6)^2 + 0.4 * (0.4)^2 ]$

=> $I = 0.0932 ML ^2$

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3 years ago