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Strike441 [17]
3 years ago
12

How many orbits/shells/energy levels does Rubidium have?

Physics
1 answer:
BabaBlast [244]3 years ago
4 0

Answer: Rubidium has 5 orbits/shells/energy levels.

Explanation: It's electron configuration is as follows: 2-8-18-8-1.

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As shown in the diagram, threee equal charges are spaced evenly in a row. The magnitude of each charge is +2e, and the distance
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Answer: 4.27 x 10^-10 N to the left

Explanation: I just took this quiz

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What is the frequency of a wave that has a wave speed of 80m/s and a wavelength of 0.20m
IrinaK [193]

speed=frequency*wavelength, so frequency=speed/wavelength. frequency=80*0.2  

Frequency = 16 Hz

6 0
3 years ago
Leta stetter hollingworth conducted pioneering work on __________. a. identity development in ethnic minorities b. cognitive pro
Elena L [17]

D.  Leta stetter hollingworth conducted pioneering work on <u>adolescent development and gifted children</u>.

<h3>Who is Leta stetter hollingworth?</h3>

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4 0
1 year ago
which wave behavior results from two waves colliding and the temporary combined wave reaults in a smaller wave than the original
choli [55]
Destructive interference is the answer
6 0
3 years ago
Scientists want to place a 3400.0 kg satellite in orbit around Mars. They plan to have the satellite orbit at a speed of 2480.0
BaLLatris [955]

Answer:

Part a)

r = 6.96 \times 10^6 m

Part b)

F_g = 3.004 \times 10^3 N

Part c)

a = 0.88 m/s^2

Part d)

v = \frac{2480}{2} = 1240 m/s

Explanation:

Part a)

As we know that the orbital speed of the satellite is given as

v = 2480 m/s

now we will have

v = \sqrt{\frac{GM_{mars}}{r}}

now we have

M_{mars} = 6.4191 \times 10^{23} kg

R_{mars} = 3.397 \times 10^6 m

now we have

2480 = \sqrt{\frac{(6.67 \times 10^{-11})(6.4191 \times 10^{23})}{r}}

r = 6.96 \times 10^6 m

Part b)

Here force between mars and satellite is the gravitational attraction force which is given as

F_g = \frac{GM_{mars} m}{r^2}

F_g = \frac{(6.67\times 10^{-11})(6.4191 \times 10^{23})(3400)}{(6.96\times 10^6)^2}

F_g = 3.004 \times 10^3 N

Part c)

Acceleration of satellite is the ratio of gravitational force and mass of the satellite

a = \frac{F_g}{m}

a = \frac{3004}{3400}

a = 0.88 m/s^2

Part d)

As we know by III law of kepler

\frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3}

here we know that T2 = 8 T1

(\frac{1}{8})^2= \frac{r_1^3}{r_2^3}

\frac{r_1}{r_2} = (\frac{1}{2})^2

so we have

r_2 = 4r_1

as we know that speed is given as

v = \sqrt{\frac{GM}{r}}

so we can say since radius is orbit becomes 4 times so the orbital speed must be half

v = \frac{2480}{2} = 1240 m/s

7 0
4 years ago
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