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katovenus [111]
3 years ago
6

Earth’s polar ice caps contain about 2.3 × 1019 kg of ice. This mass contributes essentially nothing to the moment of inertia of

the earth because it is located at the poles, close to the axis of rotation. Estimate the change in the length of the day that would be expected if the polar ice caps were to melt and the water were distributed uniformly over the surface of the earth.
Physics
1 answer:
sp2606 [1]3 years ago
3 0

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 .

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7 0
3 years ago
How can you prove the mechanical nature of sound by a simple experiment
Kruka [31]

Answer:

That is, mechanical waves cannot travel through a vacuum. This feature of mechanical waves is often demonstrated in a Physics class. A ringing bell is placed in a jar and air inside the jar is evacuated. Once air is removed from the jar, the sound of the ringing bell can no longer be heard.

6 0
3 years ago
Accomplished silver workers in india can pound silver into incredibly thin sheets, as thin as 3.00 10-7 m (about one-hundredth o
postnew [5]
The density of silver is ρ = 10500 kg/m³ approximately.

Given:
m = 1.70 kg, the mass of silver
t = 3.0 x 10⁻⁷ m, the thickness of the sheet

Let A be the area.
Then, by definition,
m = (t*A)*ρ

Therefore
A = m/(t*ρ)
    = (1.7 kg)/ [(3.0 x 10⁻⁷ m)*(10500 kg/m³)]
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Answer: 539.7 m²

8 0
3 years ago
A conducting rod moves perpendicularly through a uniform magnetic field. How does the emf change if the magnetic field is increa
Marysya12 [62]

Answer:

c) It increases by a factor of 8

Explanation:

According to Faraday's law (and Lenz' law), the induced EMF is given as the rate of change of magnetic flux.

Mathematically:

V = -dФ/dt

Magnetic flux, Ф, is given as:

Ф = BA

where B  = magnetic field strength and A = Area of object

Hence, induced EMF becomes:

V = -d(BA)/dt  or -BA/t

If the magnetic field is increased by a factor of 4, (B_n = 4B) and the time required for the rod to move is decreased by a factor of 2 (t_n = t/2), the induced EMF becomes:

V_n = -(B_nA)/t_n

V_n = \frac{-4BA}{(t/2)}\\\\V_n = \frac{-8BA}{t} \\\\V_n = 8V\\

The EMF has increased by a factor of 8.

5 0
3 years ago
If a steel cylindrical specimen is stressed nominally to 53 MPa, what stress level exists at the tip of an elliptical surface fl
ElenaW [278]

Answer:

227.9MPa

Explanation:

Length of the flaws is given by

2b = 5.8microns

b = 2.9 × 10⁻⁶m

The relation between the radius of curvature and length and width of the elliptical flaw

r = \frac{a^2}{b}

a = \sqrt{rb}

Equation for stress at the tip of an elliptical surface flaw

\sigma _t = \sigma(1 + 2\frac{b}{a}  )\\\\\sigma _t = \sigma(1 + 2\frac{b}{\sqrt{rb} })\\\\\sigma _t = \sigma(1 + 2\frac{\sqrt{b} }{\sqrt{r} })\\\\\sigma _t = \sigma(1 + 2\sqrt{\frac{b}{r} })

\sigma _t = 53 \times 10^6 (1 + 2\sqrt{\frac{2.9\times10^-^6 }{1065 \times 10^-^9} } \\\\\sigma _t = 227.9 \times 10^6\\\\\sigma _t  = 227.9MPa

4 0
3 years ago
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