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vlabodo [156]
3 years ago
8

Whats a snoogley please help

Physics
1 answer:
Ksenya-84 [330]3 years ago
5 0
Somthing that you cuddle

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(a) What is the ionization energy of a hydrogen atom that is in the n = 6 excited state? (b) For a hydrogen atom, determine the
crimeas [40]

Answer:

(a) 0.3778 eV

(b) Ratio = 0.0278

Explanation:

The Bohr's formula for the calculation of the energy of the electron in nth orbit is:

E=\frac {-13.6}{n^2}\ eV

(a) The energy of the electron in n= 6 excited state is:

E=\frac {-13.6}{6^2}\ eV

E=-0.3778\ eV

Ionisation energy is the amount of this energy required to remove the electron. Thus, |E| = 0.3778 eV

(b) For first orbit energy is:

E=\frac {-13.6}{1^2}\ eV

E=-13.6\ eV

Ratio=\frac {E_6}{E_1}

Ratio=\frac {-0.3778}{-13.6}

Ratio = 0.0278

7 0
3 years ago
A block with m = 300 grams oscillates at the end of a linear spring with k = 6.5 N/m. (assume this is a top-down view and the bl
inna [77]

Answer:

a) T=1.35s

b) amplitude = 0.0923m

Explanation:

m=300 gr

k=6.5 N/m

first we need to get the angular frequency of the motion

so we have that

ω = √(k/m)

in this case motion is a simple harmonic so the period is defined by:

T= 2π / ω

T= 2π / √(k/m)

replacing the variables...

T= 2π / √(6.5/0.3)

T=1.35s   (period of the block's motion)

and...

α max = | ω²r max |

2 = (2π/1.35)² * r max

r max= 0.0923m

6 0
3 years ago
Does mercury have more protons and electrons than tin
Klio2033 [76]
Yes it does.
On the periodic table, tin is #50 and Mercury is # 80. 
3 0
3 years ago
Please help! its really easy
Dmitry [639]

weather station - an area where weather data...

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radar - sends out signals...

5 0
3 years ago
Read 2 more answers
A person can jump a maximum horizontal distance (by using a 45◦ projectile angle) of 5 m on Earth. The acceleration of gravity i
snow_lady [41]

Answer:30 m

Explanation:

Given

Maximum Horizontal distance is 5 m on earth

launching angle=45^{\circ}

Acceleration due to gravity on earth is 9.8 m/s^2

Acceleration due to gravity on moon is \frac{9.8}{6}=1.63 m/s^2

Range of projectile is given by

R=\frac{u^2\sin 2\theta }{g}

R_{earth}=\frac{u^2\sin 2\theta }{g}=5----1

R_{moon}=\frac{u^2\sin 2\theta }{\frac{g}{6}}-----2

Divide 1 & 2

\frac{5}{R_{moon}}=\frac{1}{6}

R_{moon}=30 m

4 0
3 years ago
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