The time elapsed when the ball placed above the concave mirror and the image formed would be at the same location is 0.55 s.
<h3>Image distance</h3>
The position of the image formed is determined using the followimg mirror formula;
where;
- f is the focal length of the mirror
- v is the image distance
- u is the object distance
f = R/2
f = 1.5/2
f = 0.75 m
When the ball and its image is in the same position, u = v
The position of the ball is calculated as;
<h3>Time of motion of the ball</h3>
The time taken for the ball to travel the caluclated distance is determined as;
h = ut + ¹/₂gt²
1.5 = 0 + ¹/₂(9.8)t²
1.5 = 4.9t²
t² = 1.5/4.9
t² = 0.306
t = 0.55 s
Thus, the time elapsed when the ball and its image are at the same location is 0.55 s.
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Answer:
Explanation:
N₂ (g ) + 4H⁺( aq ) + 4e⁻ = N₂H₄ ( aq )
H⁺ ion comes from acidic medium . 4 positive charge on proton is balanced by 4 electron to make left hand side neutral because right hand side is neutral .
Answer:
The artifact is 570 years old. That is, 5.7 × 10² years.
Explanation:
Radioactive decay follows first order reaction kinetics.
Let the initial activity for fresh Carbon-14 be A₀
And the activity at any other time be A
The rate of radioactive decay is given by
dA/dt = - KA
dA/A = - kdt
Integrating the left hand side from A₀ to A₀/2 and the right hand side from 0 to t(1/2) (where t(1/2) is the radioactive isotope's half life)
In [(A₀/2)/A₀] = - k t(1/2)
In (1/2) = - k t(1/2)
- In 2 = - k t(1/2)
k = (In 2)/t₍₁,₂₎
t(1/2) is given in the question to be 5.73 × 10³ years
k = (In 2)/5730 = 0.000121 /year
dA/A = - kdt
Integrating the left hand side from A₀ to A and the right hand side from 0 to t
In (A/A₀) = - kt
A/A₀ = e⁻ᵏᵗ
A = A₀ e⁻ᵏᵗ
A = 2.8 × 10³ Bq.
A₀ = 3.0 × 10³ Bq.
2.8 × 10³ = 3.0 × 10³ e⁻ᵏᵗ
0.9333 = e⁻ᵏᵗ
e⁻ᵏᵗ = 0.9333
-kt = In 0.9333
- kt = - 0.06899
t = 0.06899/0.000121 = 570.2 years = 5.7 × 10² years
THE RADIUS OF THE TENTH ORBIT IN A HYDROGEN ATOM IS 52.9A°
<h3>How does an electron orbit work? </h3>
The three-dimensional area covering the nucleus of an atom is called electron orbital. Electrons sometimes fill low-energy orbitals which are closer to the nucleus before filling the higher ones. They mostly fill the orbitals as singly as they can and that filling is known as Hund’s rule. In the wave-like property, electrons don’t orbit a nucleus in the way a planet orbits the sun but however, but they exist as standing waves. The lowest energy possible an electron can take is the same as the fundamental frequency of a wave on a string.
the radius of the first orbit =0.0529nm
radius ∝ n²/z
radius of 10th orbit =(0.0529×100)nm=52.9A°
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