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drek231 [11]
3 years ago
13

11) An airplane is moving at a net force of zero. This means the airplane is

Physics
1 answer:
Elden [556K]3 years ago
3 0

Answer:

D. moving slow at a constant speed and traveling along a straight line

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Use the Bohr model to address this question. When a hydrogen atom makes a transition from the 5 th energy level to the 2nd, coun
iris [78.8K]

Answer:

A. 2.82 eV

B. 439nm

C. 59.5 angstroms

Explanation:

A. To calculate the energy of the photon emitted you use the following formula:

E_{n1,n2}=-13.4(\frac{1}{n_2^2}-\frac{1}{n_1^2})     (1)

n1: final state = 5

n2: initial state = 2

Where the energy is electron volts. You replace the values of n1 and n2 in the equation (1):

E_{5,2}=-13.6(\frac{1}{5^2}-\frac{1}{2^2})=2.82eV

B. The energy of the emitted photon is given by the following formula:

E=h\frac{c}{\lambda}   (2)

h: Planck's constant = 6.62*10^{-34} kgm^2/s

c: speed of light = 3*10^8 m/s

λ: wavelength of the photon

You first convert the energy from eV to J:

2.82eV*\frac{1J}{6.242*10^{18}eV}=4.517*10^{-19}J

Next, you use the equation (2) and solve for λ:

\lambda=\frac{hc}{E}=\frac{(6.62*10^{-34} kg m^2/s)(3*10^8m/s)}{4.517*10^{-19}J}=4.39*10^{-7}m=439*10^{-9}m=439nm

C. The radius of the orbit is given by:

r_n=n^2a_o   (3)

where ao is the Bohr's radius = 2.380 Angstroms

You use the equation (3) with n=5:

r_5=5^2(2.380)=59.5

hence, the radius of the atom in its 5-th state is 59.5 anstrongs

8 0
3 years ago
When is a white dwarf formed?
Anna [14]
When a red giant has insufficient mass it pretty much sheds its outer layer to leave the center core know as the white dwarf
6 0
3 years ago
Read 2 more answers
How much force is needed to accelerate an object of mass 90 kg at a rate of 1.2 m/s2
soldier1979 [14.2K]
∑F = ma = (90 kg)(1.2 m/s²) = 108 N = 100 N (1 significant digit)

3 0
3 years ago
Read 2 more answers
A motorcycle rider moving with an initial velocity of 9.2 m/s uniformly accelerates to a speed of 19.1 m/s in a distance of 32.0
Vlada [557]
Acceleration = (final velocity^2 - initial velocity^2) / 2 * distance

Acceleration = (19.1^2 - 9.2^2) / 2 * 32

Acceleration = (364.81 - 84.64) / 64

Acceleration = 280.17 / 64

Acceleration = 4.3777m/s^2

:)
6 0
3 years ago
2. FAt=p (remember p=mv). A 2.5Jg baseball moving at 8.5m/s to the left is slowed by a tipped
TiliK225 [7]

The force of the tripped catch exerted on the 2.5 Kg ball moving at 8.5 m/s to the Left is 160 N

<h3>Data obtained from the question </h3>
  • Initial velocity (u) = 8.5 m/s
  • Final velocity (v) = 7.5 m/s
  • Time (t) = 5 ms = 0.25 s
  • Mass (m) = 2.5 Kg
  • Force (F) = ?

<h3>How to determine the force</h3>

The force exerted on the ball can be obtained as follow:

F = m(v + u) / t

F = [2.5(7.5 + 8.5)]/ 0.25

F = 40 / 0.25

F = 160 N

Thus, the force exerted on the ball is 160 N

Learn more about momentum:

brainly.com/question/250648

#SPJ1

3 0
2 years ago
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