* You have an empty cylindrical metal can and two metal nuts. In your first experiment, you spin the can around its axis and place it on a rough table ( Figure P9.69 ). The can slows down due to friction, with the rotational acceleration α 1 = 30 rad/s 3 until it stops rotating in your second experiment, you first fix two nuts near the center of the can ( Figure P9.69D ) and spin it again before placing it on the table in this case, the can slows down with the rotational acceleration α 1 = 38 rad/s 2 in your third experiment. you spin the can with the nuts fixed at opposite sides of the can ( Figure P9.69c ). Now the can slows down with rotational acceleration α 3 =26 rad/s 3 . Explain (a) why α 2 > α , (b) why α 2 < α 2 , and (c) why α 3 < α .
* You have an empty cylindrical metal can and two metal nuts. In your first experiment, you spin the can around its axis and place it on a rough table ( Figure P9.69 ). The can slows down due to friction, with the rotational acceleration α 1 = 30 rad/s 3 until it stops rotating in your second experiment, you first fix two nuts near the center of the can ( Figure P9.69D ) and spin it again before placing it on the table in this case, the can slows down with the rotational acceleration α 1 = 38 rad/s 2 in your third experiment. you spin the can with the nuts fixed at opposite sides of the can ( Figure P9.69c ). Now the can slows down with rotational acceleration α 3 =26 rad/s 3 . Explain (a) why α 2 > α , (b) why α 2 < α 2 , and (c) why α 3 < α .
* You have an empty cylindrical metal can and two metal nuts. In your first experiment, you spin the can around its axis and place it on a rough table (Figure P9.69). The can slows down due to friction, with the rotational acceleration
α
1
=
30
rad/s
3
until it stops rotating in your second experiment, you first fix two nuts near the center of the can (Figure P9.69D) and spin it again before placing it on the table in this case, the can slows down with the rotational acceleration
α
1
=
38
rad/s
2
in your third experiment. you spin the can with the nuts fixed at opposite sides of the can (Figure P9.69c). Now the can slows down with rotational acceleration
α
3
=26 rad/s
3
. Explain (a) why
α
2
>
α
, (b) why
α
2
<
α
2
, and (c) why
α
3
<
α
.
For each of the actions depicted below, a magnet and/or metal loop moves with velocity v→ (v→ is constant and has the same magnitude in all parts). Determine whether a current is induced in the metal loop. If so, indicate the direction of the current in the loop, either clockwise or counterclockwise when seen from the right of the loop. The axis of the magnet is lined up with the center of the loop. For the action depicted in (Figure 5), indicate the direction of the induced current in the loop (clockwise, counterclockwise or zero, when seen from the right of the loop). I know that the current is clockwise, I just dont understand why. Please fully explain why it's clockwise, Thank you
A planar double pendulum consists of two point masses \[m_1 = 1.00~\mathrm{kg}, \qquad m_2 = 1.00~\mathrm{kg}\]connected by massless, rigid rods of lengths \[L_1 = 1.00~\mathrm{m}, \qquad L_2 = 1.20~\mathrm{m}.\]The upper rod is hinged to a fixed pivot; gravity acts vertically downward with\[g = 9.81~\mathrm{m\,s^{-2}}.\]Define the generalized coordinates \(\theta_1,\theta_2\) as the angles each rod makes with thedownward vertical (positive anticlockwise, measured in radians unless stated otherwise).At \(t=0\) the system is released from rest with \[\theta_1(0)=120^{\circ}, \qquad\theta_2(0)=-10^{\circ}, \qquad\dot{\theta}_1(0)=\dot{\theta}_2(0)=0 .\]Using the exact nonlinear equations of motion (no small-angle or planar-pendulumapproximations) and assuming the rods never stretch or slip, determine the angle\(\theta_2\) at the instant\[t = 10.0~\mathrm{s}.\]Give the result in degrees, in the interval \((-180^{\circ},180^{\circ}]\).
What are the expected readings of the ammeter and voltmeter for the circuit in the figure below? (R = 5.60 Ω, ΔV = 6.30 V)
ammeter
I =
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