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2 years ago

what is the angular speed w of the system immediately after the collision in terms f the sstem parameters and I

Physics

1 answer:

denis23 [38]2 years ago

5 0

Answer: hello some part of your question is missing attached below is the missing detail

answer :

wf = M( v cos∅ )D / I

Explanation:

The Angular speed wf of the system after collision in terms of the system parameters and I can be expressed as

considering angular momentum conservation

Li = Lf

M( v cos∅ ) D = ( ML^2 / 3 + mD^2 ) wf

where ; ( ML^2 / 3 + mD^2 ) = I ( Inertia )

In terms of system parameters and I

wf = M( v cos∅ )D / I

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3 years ago

Suppose you want to determine the resistance of a resistor that is nominally 100 . You should be able to apply 10 V across the r

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Answer:

a) For y = 102 mA, R = 98.039 ohms

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b) Check explanatios for b

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$R = \frac{10}{0.097} \\R = 103 .09 ohms$

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$y = \left[\begin{array}{ccc}y_{1} \\y_{2} \end{array}\right]$

$y = \left[\begin{array}{ccc}102*10^{-3} \\97*10^{-3} \end{array}\right]$

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$x = \left[\begin{array}{ccc}100\end{array}\right]$

$\left[\begin{array}{ccc}102*10^{-3} \\97*10^{-3} \end{array}\right] = \left[\begin{array}{ccc}G_{1} \\G_{2} \end{array}\right] \left[\begin{array}{ccc}100\end{array}\right]\\\left[\begin{array}{ccc}G_{1} \\G_{2} \end{array}\right] = \left[\begin{array}{ccc}102*10^{-5} \\97*10^{-5} \end{array}\right]$

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3 years ago

Read 2 more answers

A person holds a ladder horizontally at its center. Treating the ladder as a uniform rod of length 4.15 m and mass 7.98 kg, find

Ludmilka [50]

Answer:

4.535 N.m

Explanation:

To solve this question, we're going to use the formula for moment of inertia

I = mL²/12

Where

I = moment of inertia

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On solving we have

I = 7.98 * (4.15)² / 12

I = (7.98 * 17.2225) / 12

I = 137.44 / 12

I = 11.45 kg·m²

That is the moment of inertia about the center.

Using this moment of inertia, we multiply it by the angular acceleration to get the needed torque. So that

τ = 11.453 kg·m² * 0.395 rad/s²

τ = 4.535 N·m

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3 years ago