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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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Answer:

a. Quadruped arm and opposite leg raise

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This method is usually used as an alternative to iso-abs exercise or also known as a bridge, which allows you to exercise the abdominal and spinal area at the same time.

It is also used together with other exercises for the treatment of hyperlordosis.

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8.3 cm^2

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1.Which of the following is evidence that the Earth is round?
PtichkaEL [24]

Answer:

1\D. Both A and C

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5 0
3 years ago
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Ice skaters often end their performances with spin turns, where they spin very fast about their center of mass with their arms f
vampirchik [111]

Answer:

\large \boxed{\text{30 rev/s}}

Explanation:

This question is based on the Law of Conservation of Angular Momentum.

Angular momentum (L) equals the moment of inertia (I) times the angular speed (ω).

L = Iω

If momentum is conserved,

I₁ω₁ = I₂ω₂

Data:

 I₁ = 3.5    kg·m²s⁻¹

ω₁ = 6.0    rev·s⁻¹

 I₂ = 0.70 kg·m²s⁻¹

Calculation:

\begin{array}{rcl}I_{1}\omega_{1} &= &I_{2}\omega_{2}\\\text{3.5 kg$\cdot$m$^{2}$}\times \text{6.0 rev/s} &= &\text{0.70 kg$\cdot$m$^{2}$}\times\omega_{2}\\\text{21 rev/s} &= &0.70\omega_{2}\\\omega_{2} & = & \dfrac{\text{21 rev/s}}{0.70}\\\\&=&\textbf{30 rev/s}\\\end{array}\\\text{The skater's final rotational speed is $\large \boxed{\textbf{30 rev/s}}$}

8 0
3 years ago
RP 1 - Specific Heat Capacity GCSE Exam Questions (B.S.G)
Stels [109]

When the temperature of 0.50 kg of water decreases by 22 °C, the energy transferred to the surroundings from the water is -46.2 kJ.

A sample of 0.50 kg of water boils (reaches 100 °C). After a while, its temperature decreases by 22 °C.

We can calculate the energy transferred to the surroundings from the water in the form of heat (Q) using the following expression.

Q = c \times m \times \Delta T = \frac{4200J}{kg.\° C}  \times 0.50kg \times (-22\° C) \times \frac{1kJ}{1000J} = -46.2 kJ

where,

  • c: specific heat capacity of water
  • m: mass of water
  • ΔT: change in the temperature

When the temperature of 0.50 kg of water decreases by 22 °C, the energy transferred to the surroundings from the water is -46.2 kJ.

Learn more: brainly.com/question/16104165

8 0
2 years ago
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