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kykrilka [37]
3 years ago
8

e force acting between two charged particles A and B is 5.2 × 10-5 newtons. Charges A and B are 2.4 × 10-2 meters apart. If the

charge on particle A is 7.2 × 10-8 coulombs, what is the charge of particle B? (k = 9.0 × 109 newton·meters2/coulomb2)
Physics
2 answers:
anastassius [24]3 years ago
7 0
The force acting between the particles is

F=k \frac{Q_{1}Q_{2}}{r^2}
Then
Q_{2}= \frac{5.2 \times 10^-^5 \times 0.024^2}{ 9.0 \times 10^9 7.2 \times 10^-^8} =4.622 \times 10^-^1^1C




True [87]3 years ago
6 0

Answer:

the answer is b for plato users

Explanation:

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kifflom [539]

Answer:

Exophthalmos

Explanation:

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The probability that a battery will last 10 hr or more is 0.8, and the probability that it will last 15 hr or more is 0.11. Give
podryga [215]
We are given with the data that says the probability that a battery will last 10 hr or more is 0.8 and the probability <span>that it will last 15 hr or more is 0.11. In this case, the probability that the battery lasts at least 10 hours and even 15 hrs more is 0.11 / 0.8 or equal to 13.75 percent.</span>
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Intelligence test. If you get it right ,you are a critical thinker. You were in the garden,there are 34 people in the yard. You
bagirrra123 [75]

The riddle is solved by knowing that there is only one person left in the garden and that is you.

<h3>How many people are in the garden?</h3>

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Learn more about solving a riddle:brainly.com/question/25063408

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8 0
2 years ago
When you look at yourself in the mirror, what is the approximate angle of incidence of the light rays?
Kitty [74]
46 degrees
I hope this helps :)
3 0
3 years ago
Calculate the energy, wavelength, and frequency of the emitted photon when an electron moves from an energy level of -3.40 eV to
jasenka [17]

Answer:

(a) The energy of the photon is 1.632 x 10^{-8} J.

(b) The wavelength of the photon is 1.2 x 10^{-17} m.

(c) The frequency of the photon is 2.47 x 10^{25} Hz.

Explanation:

Let;

E_{1} = -13.60 ev

E_{2} = -3.40 ev

(a) Energy of the emitted photon can be determined as;

E_{2} - E_{1} = -3.40 - (-13.60)

           = -3.40 + 13.60

           = 10.20 eV

           = 10.20(1.6 x 10^{-9})

E_{2} - E_{1} = 1.632 x 10^{-8} Joules

The energy of the emitted photon is 10.20 eV (or 1.632 x 10^{-8} Joules).

(b) The wavelength, λ, can be determined as;

E = (hc)/ λ

where: E is the energy of the photon, h is the Planck's constant (6.6 x 10^{-34} Js), c is the speed of light (3 x 10^{8} m/s) and λ is the wavelength.

10.20(1.6 x 10^{-9}) = (6.6 x 10^{-34} * 3 x 10^{8})/ λ

λ = \frac{1.98*10^{-25} }{1.632*10^{-8} }

  = 1.213 x 10^{-17}

Wavelength of the photon is 1.2 x 10^{-17} m.

(c) The frequency can be determined by;

E = hf

where f is the frequency of the photon.

1.632 x 10^{-8}  = 6.6 x 10^{-34} x f

f = \frac{1.632*10^{-8} }{6.6*10^{-34} }

 = 2.47 x 10^{25} Hz

Frequency of the emitted photon is 2.47 x 10^{25} Hz.

6 0
3 years ago
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