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Mice21 [21]
3 years ago
8

What is the answer to these questions?

Physics
1 answer:
Oksana_A [137]3 years ago
7 0

Answer:

THES IS NOT

Explanation:

THIS PAPAER IS A FAKE PAPAER BEACISE POGI TALAGA AKO

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1 Calculate the size of the quantum involved in the excitation of (a) an electronic motion of frequency 1.0 × 1015 Hz, (b) a mol
torisob [31]

Answer: a) E= 6.63x10^-19J

E= 3.97×10^2KJ/mol

b) E = 3.31×10^-19J

E= 18.8×10^4 KJ/mol

C) E = 1.32×10^-33J

E= 8.01×10^-10KJ/mol

Explanation:

a) E = h ×f

h= planks constant= 6.626×10^-34

E=(6.626×10^-34)×(1.0×10^15)

E=6.63×10^-19J

1mole =6.02×10^23

E=( 6.63×10^-19)×(6.02×10^23)

E=3.97×10^2KJ/mol

b) E =(6.626×10^-34)/(1.0×10^15)

E=3.13×10^-19J

E= 3.13×10^-19) ×(6.02×10^23)

E= 18.8×10^3KJ/MOL

c) E= (6.626×10^-34) /0.5

E= 1.33×10^-33J

E= (1.33×10^-33) ×(6.02×10^23)

E= 8.01×10^-10KJ/mol

8 0
3 years ago
Find the magnitude of the sum of two vectors; A is 5 km , and B is 7 , when the angle btween them is 120
Dimas [21]

Answer: 6.24 km

Explanation:

Given

The magnitude of the first vector(say) \left |  a\right |=5\ km

the magnitude of the second vector(say) \left |  b\right |=7\ km

the angle between them is 120^{\circ}

The resultant vector magnitude is given by

\left |  \vec{R}\right |=\sqrt{a^2+b^2+2ab\cos \theta}

\left |  \vec{R}\right |=\sqrt{5^2+7^2+2\times 5\times 7\cdot \cos 120^{\circ}}\\\left |  \vec{R}\right |=\sqrt{74-35}=\sqrt{39}\\\left |  \vec{R}\right |=6.24\ km

4 0
3 years ago
The apparent reversal of motion of planets against the backdrop of fixed stars is called
Free_Kalibri [48]

The planets move eastward against the background of fixed stars with the exception of Venus, Uranus and Pluto moving westward as seen in Earth's sky . This apparent retreating movement is called Retrograde motion. It is illusion created by Earth's movement going by outer planets in their respective orbits.

7 0
3 years ago
Read 2 more answers
Who is responsible for developing the three laws of planetary motion?
Ivenika [448]
Johannes Kepler- he did it by observing the ‘Tycho Brahe’. His 3rd law was published 10 years later to his first two laws.
3 0
3 years ago
A bullet is fired into the air at an angle of 45°. How far does it travel before it is 1,000 feet above the ground? (Assume that
Readme [11.4K]

Answer:

It travels 1414 feets.

Explanation:

Let's take the length the bullet travels <em>l </em>as the hypotenuse of a right triangle and the height it reaches one of its sides. Since we got the angle α at which it was fired and the height <em>h</em> it reached, we can calculate <em>l</em> using the <em>sin(α)</em> function:

sin(\alpha )=\frac{opposite side}{hypotenuse}\\sin(\alpha)=\frac{h}{l}\\l=\frac{h}{sin(\alpha)}

Replacing:

l=\frac{1000ft}{sin(\frac{\pi}{4})}

Solving and roundin to the nearest foot:

l=1414 ft

3 0
3 years ago
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