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Free_Kalibri [48]
3 years ago
9

In this problem, you will practice applying this formula to several situations involving angular acceleration. in all of these s

ituations, two objects of masses m1 and m2 are attached to a seesaw. the seesaw is made of a bar that has length l and is pivoted so that it is free to rotate in the vertical plane without friction.assume that the pivot is attached tot he center of the bar. you are to find the angular acceleration of the seesaw when it is set in motion from the horizontal position. in all cases, assume that m1>m2.
Physics
1 answer:
MatroZZZ [7]3 years ago
6 0
<span>A- 2*[(m1 - m2)/(m1 + m2)]*g/L The rotation is in the counterclockwise direction and the angular acceleration is positive. B- 2*[(m1 - m2)/(m1+ m2 +mbar/3)]*g/L The rotation is in the counterclockwise direction and the angular acceleration is positive.</span>
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Ok I have no clue for this one I’m not sure what to make out of this one please help
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8 0
3 years ago
A rowboat has a momentum of 241.5 kg*m/s. if the rowboat has a mass of 115kg, what its it's speed?
Lera25 [3.4K]
Momentum = mass*velocity(or speed)

241.5 kg*m/s = 115 kg * v
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3 years ago
If you throw a baseball straight up,What is its velocity at the highest point?​
sveta [45]

Answer: 52.

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7 0
3 years ago
Read 2 more answers
(a) What is the ionization energy of a hydrogen atom that is in the n = 6 excited state? (b) For a hydrogen atom, determine the
crimeas [40]

Answer:

(a) 0.3778 eV

(b) Ratio = 0.0278

Explanation:

The Bohr's formula for the calculation of the energy of the electron in nth orbit is:

E=\frac {-13.6}{n^2}\ eV

(a) The energy of the electron in n= 6 excited state is:

E=\frac {-13.6}{6^2}\ eV

E=-0.3778\ eV

Ionisation energy is the amount of this energy required to remove the electron. Thus, |E| = 0.3778 eV

(b) For first orbit energy is:

E=\frac {-13.6}{1^2}\ eV

E=-13.6\ eV

Ratio=\frac {E_6}{E_1}

Ratio=\frac {-0.3778}{-13.6}

Ratio = 0.0278

7 0
4 years ago
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