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AURORKA [14]
3 years ago
15

An astronaut with a mass of 91 kg is 0.30 m above the moons surface. The astronauts potential energy is 46 J. Calculate the free

-fall acceleration on the moon?
Physics
1 answer:
Blababa [14]3 years ago
3 0

Answer:

the free-fall acceleration on the moon is 1.68 m/s^2

Explanation:

recall the formula for the gravitational potential energy (under acceleration of gravity "g"):

PE = m * g * h

replacing with our values for the problem:

46 J = 91 * g * 0.3

solve for the "g" on the Moon:

g = 46 / (91 * 0.3)

g = 1.68  m/s^2

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When the hydrogen atom makes the transition from the n=2 to the n=1 energy level, it emits a photon. This photon can be absorbed
bazaltina [42]

Answer:

n_{fn}= 4

Explanation:

To solve this exercise we will use Bohr's atomic model

               E_{n} = - 13.606 / n²     [eV]

The transition from level n = 2 to level n = 1 is valid

               E_{21} = - 13.606 [¼ -1/1]

               E_{21} = 10.2045 eV

Bohr's model for atoms with only one electron is

               E_{n} = -13.606 Z² / n²

Where Z is the atomic number of the atom.

In this case the helium atom has an atomic number of Z = 2 from the level n₀ = 2 let's look up to what level it reaches

         ΔE = -13.606 [4 /  n_{fn}² - 4/4]

         4 / n_{fn}² = -ΔE / 13.606 + 1

         4 / n_{fn}² = -10.2045 / 13.606 +1 = -0.75 +1

         4 / n_{fn}² = 0.25

        n_{fn} = √ 4 / 0.25

        n_{fn}= 4

8 0
3 years ago
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3 0
3 years ago
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What constant acceleration, in SI units, must a car have to go from zero to 60 mph in 10 s? How far has the car traveled when it
nalin [4]

Answer:

Explanation:

initial velocity, u = 0

final velocity, v = 60 mph = 26.8 m/s

time t = 10 s

Let a be the acceleration and s be he distance traveled.

Use first equation of motion

v = u + a t

26.8 = 0 +  a x 10

a = 2.68 m/s

Use second equation of motion

s = ut + 1/2 at²

s = 0 + 0.5 x 2.68 x 10 x 10

s = 134 m

As, 1 m = 3.28 ft

So, s = 134 x 3.28 ft

s = 439.6 ft

7 0
3 years ago
What is the displacement from a starting position of (14.0, 3.0) m to a final position of (−3.0, −4.0) m?
dem82 [27]

Answer:

change in y = -7

change in x = -17

magnitude of displacement = sqrt(7^2+17^2)

tan of angle below -x axis = 7/17

because in third quadrant where x and y are negative

3 0
3 years ago
. Mass 1 has a momentum of 20 kg*m/s. Mass 2 has a momentum of 50 kg*m/s.
sesenic [268]

Answer:

55kgm/s

Explanation:

3 0
3 years ago
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