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2 years ago

All the actinoids, atomic number 89 through 103 have seven electron orbits. True False

1 answer:

Pie2 years ago

7 0

Answer is: true.
Actinoids are in 7 period in Periodic table and valence orbit is 7s².
Ac - [Rn] 6d¹ 7s².
Th - [Rn] 6d² 7s².
Pa - [Rn] 5f² 6d¹ 7s².
U - [Rn] 5f³ 6d¹ 7s².
Np - [Rn] 5f⁴ 6d¹ 7s².
Pu - [Rn] 5f⁶ 7s².
Am - [Rn] 5f⁷ 7s².
Cm - [Rn] 5f⁷ 6d¹ 7s².
Bk - [Rn] 5f⁹ 7s².
Cf - [Rn] 5f¹⁰ 7s².

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If the planes of a crystal are 3.50 (1 A= 10^-10m = Ångstrom unit) apart, what wavelength of electromagnetic waves are needed so

goldenfox [79]

Answer:

λ = 2.62 x 10⁻¹⁰ m = 0.262 nm

Explanation:

We can use Bragg's Law's equation to solve this problem. The Bragg's Law's equation is written as follows:

mλ = 2d Sin θ

where,

m = order of reflection = 1

λ = wavelength = ?

d = distance between the planes of crystal = 3.5 x 10⁻¹⁰ m

θ = strike angle of waves on plane = 22°

Therefore, substituting the respective values in the equation, we get:

(1)λ = (2)(3.5 x 10⁻¹⁰ m)(Sin 22°)

λ = 2.62 x 10⁻¹⁰ m = 0.262 nm

4 0

2 years ago

A ringing bell sends sound waves in all directions places sides

Strike441 [17]

Answer:

direction

Explanation:

because particles surround the bell, so when the bell vibrates, it causes particles surrounding it to vibrate back and forth vigorously. as these particles vibrate they collide with the neighbouring particles, passing on the energy.

hope this is what you are asking, if not please report it so that someone else gets to try it.

4 0

2 years ago

A person walks 5.0kilometers north, then 5.0 kilometers east. His displacement is closest to ? A. 10 kilometers northwest B. 7.1

Rainbow [258]

Use vector analysis and calculate resultant vector using Pythagoras theorem:
5^2 + 5^2 = 50
Square root of 50 = approx 7.1 km NE
Therefore the answer is D

4 0

3 years ago

Six artificial satellites circle a space station at constant speed. The mass m of each satellite, distance L from the space stat

nikklg [1K]

Answers:

a) $T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}$

b) $a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}$

Explanation:

a) Since we are told the satellites circle the space station at constant speed, we can assume they follow a uniform circular motion and their tangential speeds $V$ are given by:

$V=\omega L=\frac{2\pi}{T} L$ (1)

Where:

$\omega$ is the angular frequency

$L$ is the radius of the orbit of each satellite

$T$ is the period of the orbit of each satellite

Isolating $T$ :

$T=\frac{2 \pi L}{V}$ (2)

Applying this equation to each satellite:

$T_{1}=\frac{2 \pi L}{V_{1}}=261.79 s$ (3)

$T_{2}=\frac{2 \pi L}{V_{2}}=1570.79 s$ (4)

$T_{3}=\frac{2 \pi L}{V_{3}}=196.349 s$ (5)

$T_{4}=\frac{2 \pi L}{V_{4}}=98.174 s$ (6)

$T_{5}=\frac{2 \pi L}{V_{5}}=785.398 s$ (7)

$T_{6}=\frac{2 \pi L}{V_{6}}=196.349 s$ (8)

Ordering this periods from largest to smallest:

$T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}$

b) Acceleration $a$ is defined as the variation of velocity in time:

$a=\frac{V}{T}$ (9)

Applying this equation to each satellite:

$a_{1}=\frac{V_{1}}{T_{1}}=0.458 m/s^{2}$ (10)

$a_{2}=\frac{V_{2}}{T_{2}}=0.0254 m/s^{2}$ (11)

$a_{3}=\frac{V_{3}}{T_{3}}=0.4074 m/s^{2}$ (12)

$a_{4}=\frac{V_{4}}{T_{4}}=1.629 m/s^{2}$ (13)

$a_{5}=\frac{V_{5}}{T_{5}}=0.101 m/s^{2}$ (14)

$a_{6}=\frac{V_{6}}{T_{6}}=0.814 m/s^{2}$ (15)

Ordering this acceerations from largest to smallest:

$a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}$

4 0

2 years ago

An object with a mass m slides down a rough 370 inclined plane where the coefficient of kinetic friction is 0.20. If the plane i

Svetllana [295]

Answer:

$v \approx 9.312\,\frac{m}{s}$

Explanation:

The Free Body Diagram of the system is presented in the image attached below. The final speed is determined by means of the Principle of Energy Conservation and the Work-Energy Theorem:

$K_{A} + U_{g,A} = K_{B} + U_{g,B} + W_{loss}$