Answer:
Explanation:
Initial moment of inertia of the earth I₁ = 2/5 MR² , M is mss of the earth and R is the radius . If ice melts , it forms an equivalent shell of mass 2.3 x 10¹⁹ Kg
Final moment of inertia I₂ = 2/5 M R² + 2/3 x 2.3 x 10¹⁹ x R²
For change in period of rotation we shall apply conservation of angular momentum law
I₁ ω₁ = I₂ ω₂ , ω₁ and ω₂ are angular velocities initially and finally .
I₁ / I₂ = ω₂ / ω₁
I₁ / I₂ = T₁ / T₂ , T₁ , T₂ are time period initially and finally .
T₂ / T₁ = I₂ / I₁
(2/5 M R² + 2/3 x 2.3 x 10¹⁹ x R²) / 2/5 MR²
1 + 5 / 3 x 2.3 x 10¹⁹ / M
= 1 + 5 / 3 x 2.3 x 10¹⁹ / 5.97 x 10²⁴
= 1 + .0000064
T₂ = 24 (1 + .0000064)
= 24 hours + .55 s
change in length of the day = .55 s .
Direction. Velocity is a vector that describes both speed and direction, while speed is a scalar that describes only speed regardless of direction.
Answer:
the number density of the protons in the beam is 3.2 × 10¹³ m⁻³
Explanation:
Given that;
diameter D = 2.0 mm
current I = 1.0 mA
K.E of each proton is 20 MeV
the number density of the protons in the beam = ?
Now, we make use of the relation between current and drift velocity
I = MeAv ⇒ 1 / eAv
The kinetic energy of protons is given by;
K = 
v²
v = √( 2K / )
lets relate the cross-sectional area A of the beam to its diameter D;
A = 
πD²
now, we substitute for v and A
n = I / πeD² ×√( 2K / )
n = 4I/π eD² × √( / 2K )
so we plug in our values;
n = ((4×1.0 mA)/(π(1.602×10⁻¹⁹C)(2mm)²) × √(1.673×10⁻²⁷kg / 2×( 20 MeV)(1.602×10⁻¹⁹ J/ev )
n = 1.98695 × 10¹⁸ × 1.6157967 × 10⁻⁵
n = 3.2 × 10¹³ m⁻³
Therefore, the number density of the protons in the beam is 3.2 × 10¹³ m⁻³
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