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Viefleur [7K]
4 years ago
15

Calculate the magnitude of the maximum orbital angular momentum Lmax for an electron in a hydrogen atom for states with a princi

pal quantum number of 1.
Physics
1 answer:
erica [24]4 years ago
3 0

Answer:

L_max= 0

Explanation:

The formula for magnitude of maximum orbital angular momentum is given by

L_{max}= \sqrt{l(l+1)\bar{h}}

l= orbital quantum number

l= n-1

n= shell number or principal quantum number

for n=1 , l=0

therefore, L_{max}= \sqrt{0(0+1)\bar{h}}

L_max= 0

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A 0.500 kg bullet is fired from a gun at 25.0 m/s, how much kinetic energy does it have?
slamgirl [31]

Considering the definition of kinetic energy, the bullet has a kinetic energy of 156.25 J.

<h3>Kinetic energy</h3>

Kinetic energy is a form of energy. It is defined as the energy associated with bodies that are in motion and this energy depends on the mass and speed of the body.

Kinetic energy is defined as the amount of work necessary to accelerate a body of a given mass and in a rest position, until it reaches a given speed. Once this point is reached, the amount of accumulated kinetic energy will remain the same unless there is a change in speed or the body returns to its rest state by applying a force to it.

The kinetic energy is represented by the following expression:

Ec= ½ mv²

Where:

  • Ec is the kinetic energy, which is measured in Joules (J).
  • m is the mass measured in kilograms (kg).
  • v is the speed measured in meters over seconds (m/s).

<h3>Kinetic energy of a bullet</h3>

In this case, you know:

  • m= 0.500 kg
  • v= 25 m/s

Replacing in the definition of kinetic energy:

Ec= ½ ×0.500 kg× (25 m/s)²

Solving:

<u><em>Ec= 156.25 J</em></u>

Finally, the bullet has a kinetic energy of 156.25 J.

Learn more about kinetic energy:

brainly.com/question/25959744

brainly.com/question/14028892

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2 years ago
HELP PLS!<br> (LOOK AT THE PICTURE)
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Using energy considerations, calculate the average force (in N) a 67.0 kg sprinter exerts backward on the track to accelerate fr
Illusion [34]

Answer:

F_{sprinter}=110.4N

Explanation:

Given data

Mass m=67.0 kg

Final Speed vf=8.00 m/s

Initial Speed vi=2.00 m/s

Distance d=25.0 m

Force F=30.0 N

From work-energy theorem we know that the work done equals the change in kinetic energy

W=ΔK=Kf-Ki=1/2mvf²-1/2mvi²

And

W=F_{total}.d

So

W=1/2mv_{f}^2-1/2mv_{i}^2\\F_{total}=\frac{1/2mv_{f}^2-1/2mv_{i}^2}{d} \\F_{total}=\frac{1/2(67.0kg)(8.00m/s)^2-1/2(67.0kg)(2.00m/s)^2}{25.0m} \\F_{total}=80.4N

and we know that the force the sprinter exerted Fsprinter the force of the headwind Fwind=30.0N

So

F_{sprinter}=F_{total}+F_{wind}\\F_{sprinter}=80.4N+30N\\F_{sprinter}=110.4N  

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3 years ago
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--  The student's distance traveled is 200 meters.

--  The student's displacement is 141 meters Northeast.

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I think the blank would be "dominant".
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