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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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Therefore, if the block moves from its position of maximum spring stretch to maximum spring compression in 0.25 s, the time required for a full cycle is twice as much; T = 0.5 s.

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Students perform a set of experiments by placing a block of mass m against a spring, compressing the spring a distance x along a
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Increasing the angle of inclination of the plane decreases the velocity of the block as it leaves the spring.

  • The statement that indicates how the relationship between <em>v</em> and <em>x</em> changes is;<u> As </u><u><em>x</em></u><u> increases, </u><u><em>v</em></u><u> increases, but the relationship is no longer linear and the values of </u><u><em>v</em></u><u> will be less for the same value of </u><u><em>x</em></u><u>.</u>

Reasons:

The energy given  to the block by the spring = \mathbf{0.5  \cdot k  \cdot x^2}

According to the principle of conservation of energy, we have;

On a flat plane, energy given to the block = 0.5  \cdot k  \cdot x^2 = kinetic energy of

block = 0.5  \cdot m  \cdot v^2

Therefore;

0.5·k·x² = 0.5·m·v²

Which gives;

x² ∝ v²

x ∝ v

On a plane inclined at an angle θ, we have;

The energy of the spring = \mathbf{0.5  \cdot k  \cdot x^2}

  • The force of the weight of the block on the string, F = m \cdot g  \cdot sin(\theta)

The energy given to the block = 0.5 \cdot k \cdot x^2 - m \cdot g  \cdot sin(\theta) = The kinetic energy of block as it leaves the spring = \mathbf{0.5  \cdot m  \cdot v^2}

Which gives;

0.5 \cdot k \cdot x^2 - m \cdot g  \cdot sin(\theta) = 0.5  \cdot m  \cdot v^2

Which is of the form;

a·x² - b = c·v²

a·x² + c·v² = b

Where;

a, b, and <em>c</em> are constants

The graph of the equation a·x² + c·v² = b  is an ellipse

Therefore;

  • As <em>x</em> increases, <em>v</em> increases, however, the value of <em>v</em> obtained will be lesser than the same value of <em>x</em> as when the block is on a flat plane.

<em>Please find attached a drawing related to the question obtained from a similar question online</em>

<em>The possible question options are;</em>

  • <em>As x increases, v increases, but the relationship is no longer linear and the values of v will be less for the same value of x</em>
  • <em>The relationship is no longer linear and v will be more for the same value of x</em>
  • <em>The relationship is still linear, with lesser value of v</em>
  • <em>The relationship is still linear, with higher value of v</em>
  • <em>The relationship is still linear, but vary inversely, such that as x increases, v decreases</em>

<em />

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An electric field of magnitude 2.35 V/m is oriented at an angle of 25.0° with respect to the positive z-direction. Determine the
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Answer:

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Explanation:

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We need to calculate the flux

Using formula of the magnetic flux

\phi=E\cdot A

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Put the value into the formula

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