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goldenfox [79]
3 years ago
9

A thin, uniform rod is hinged at its midpoint. to begin with, one-half of the rod is bent upward and is perpendicular to the oth

er half. this bent object is rotating at an angular velocity of 6.9 rad/s about an axis that is perpendicular to the left end of the rod and parallel to the rod's upward half (see the drawing). without the aid of external torques, the rod suddenly assumes its straight shape. what is the angular velocity of the straight rod?
Physics
1 answer:
bazaltina [42]3 years ago
5 0
<span>Answer: initial I = (m/2)L²/3 + (m/2)L² where L = ½ the length of the rod, and the vertical half can be treated as a point mass. initial I = mL²(1/6 + 1/2) = 2mL²/3 final I = m(2L)²/3 = 4mL²/3 Since I has doubled and momentum is conserved, ω has halved. ω = 3.9 rad/s. Formulaically: 2mL²/3 * 7.8rad/s = 4mL²/3 * ω</span>
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Calculate the separation between the two lowest levels for an O2 molecule in a one-dimensional container of length 5.0 cm. At wh
MariettaO [177]

Answer:

The separation between the two lowest levels =  1.24 * 10^{-39}J

The values of n where the energy of molecule reaches 1/2 kT at 300K = 2.2 * 10^{9}

The separation at this level = 1.8 * 10^{-30}J

Explanation:

Knowing the formula

En = \frac{n^{2} h^{2}  }{8 mL^{2} }

Mass of oxygen molecule

m (O2) = 32 amu * \frac{1.6605 * 10^{-27 kg} }{1 amu}

So the energy diference between the two lowest levels:

E2 - E1 = \frac{3h^{2} }{8mL^{2} }

E2 - E1 =  \frac{3 * (6.626 * 10^{-34} Js)^{2} } {8 * 32 amu * (\frac{1.6605 * 10^{-27 kg} }{1 amu})* (5*10^{-2})^{2}   } = 1.24 * 10^{-39}J

Now we should find n where the energy of molecule reaches 1/2 kT

En = \frac{n^{2} h^{2}  }{8 mL^{2} } = \frac{1}{2}kT

\frac{h^{2}  }{8 mL^{2} } = 4.13 * 10^{-14}J

n^{2} *  (4.13 * 10^{-14}J) = \frac{1}{2} (1.38 * 10^{-23}JK^{-1}) * 300K

n = 2.2 * 10^{9}

by the end is necessary to calculate the separation of the level

En - En-1 = (n^{2} - (n - 1)^{2}) * \frac{h^{2}  }{8 mL^{2} }

              = 1.8 * 10^{-30}J

4 0
3 years ago
After 24.0 days 2.00 milligrams of an original 128.0. Milligram sample remain what is the half life of the sample
alina1380 [7]

Answer:

4 days

either multiply 128 by .5 until you get to 2 counting each time or use 2 formulas ln(n2/n1)=-k(t2-t1) to get k then input k into ln(2)=k*t1/2

n2 is final amount and n1 is beginning and t is either time elapsed as in the first formula or the actual half life that is t1/2

Explanation:

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3 years ago
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Nataly_w [17]
Volume doesn't depend on what the substance is, only on how much of it there is.
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3 years ago
What causes an impact crater to form?
Scrat [10]

Answer:

it is the impact of a meteorite, volcanic activity, or an explosion.

4 0
3 years ago
The tip of the second hand of a clock moves in a circle of 20 cm circumference. In one minute the hand makes a complete revoluti
Cerrena [4.2K]

Answer:

v_{avg} = 0

Explanation:

As we know that average velocity is defined as the ratio of total displacement of the object and its time interval.

so here we can say

v_{avg} = \frac{displacement}{time}

now we know that in one complete revolution the total displacement of the tip of the seconds hand is zero

because it will have same position after one complete revolution from where it starts

so here we can say that the average velocity will be zero

v_{avg} = 0

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3 years ago
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