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Mamont248 [21]
3 years ago
11

A skater is spinning with his arms outstretched. He has a 2 lb weight in each hand. In an attempt to change his angular velocity

he lets go of both weights (by just opening his grip). Does he succeed in changing his angular velocity
Physics
1 answer:
Softa [21]3 years ago
8 0

Answer:

No

Explanation:

Changing momentum of any kind requires work. Work is a force acting over a distance. While holding the weights at arms length and spinning will create a force (centripetal), there is no radial distance change incurred. Releasing the weights will reduce the force to zero, still no work done and no change in angular momentum.

If he was holding the weights at arms length while spinning and he pull his hands to his chest, there now exists both the centripetal force and a distance in the direction of that force (inward radial) this work will result in an increase in angular velocity as moment of inertia has decreased with the work done.

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Answer:

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Explanation:

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3 years ago
The small metal cylinder has a mass of 0.20 kgkg, the coefficient of static friction between the cylinder and the turntable is 0
deff fn [24]

Answer:

velocity of the metal cylinder = 0.343 m/s

Explanation:

Force putting the metal cylinder is given by

F = mv²/r

But this force will balance the frictional force between the metal cylinder and the turntable

The frictional force is given by

μN = μ × mg = 0.08 × 0.2 × 9.81 = 0.15696 N

r = 0.15 m, m = 0.2 kg,

F = mv²/r = 0.2 v²/(0.15) = 1.3333 v²

1.3333 v² = 0.15696

v² = 0.117

v = 0.343 m/s

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3 years ago
is it true that the composition of the atmosphere changes every few kilometers as you move away from earth
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3 years ago
Three identical resistors are connected in parallel. The equivalent resistance increases by 630 when one resistor is removed and
strojnjashka [21]

Answer:

each resistor is 540 Ω

Explanation:

Let's assign the letter R to the resistance of the three resistors involved in this problem. So, to start with, the three resistors are placed in parallel, which results in an equivalent resistance R_e defined by the formula:

\frac{1}{R_e}=\frac{1}{R} } +\frac{1}{R} } +\frac{1}{R} \\\frac{1}{R_e}=\frac{3}{R} \\R_e=\frac{R}{3}

Therefore, R/3 is the equivalent resistance of the initial circuit.

In the second circuit, two of the resistors are in parallel, so they are equivalent to:

\frac{1}{R'_e}=\frac{1}{R} +\frac{1}{R}\\\frac{1}{R'_e}=\frac{2}{R} \\R'_e=\frac{R}{2} \\

and when this is combined with the third resistor in series, the equivalent resistance (R''_e) of this new circuit becomes the addition of the above calculated resistance plus the resistor R (because these are connected in series):

R''_e=R'_e+R\\R''_e=\frac{R}{2} +R\\R''_e=\frac{3R}{2}

The problem states that the difference between the equivalent resistances in both circuits is given by:

R''_e=R_e+630 \,\Omega

so, we can replace our found values for the equivalent resistors (which are both in terms of R) and solve for R in this last equation:

\frac{3R}{2} =\frac{R}{3} +630\,\Omega\\\frac{3R}{2} -\frac{R}{3} = 630\,\Omega\\\frac{7R}{6} = 630\,\Omega\\\\R=\frac{6}{7} *630\,\Omega\\R=540\,\Omega

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