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stealth61 [152]

3 years ago

11

At ground level g is 9.8m/s^2. Suppose the earth started to increase its angular velocity. How long would a day be when people o

n the equator were just 'thrown off'? Why is the expression 'thrown off' a bad one ?

2 answers:

Elina [12.6K]3 years ago

6 0

Hello there.
<span>
At ground level g is 9.8m/s^2. Suppose the earth started to increase its angular velocity. How long would a day be when people on the equator were just 'thrown off'? Why is the expression 'thrown off' a bad one ?
</span>
People would not be thrown off.

velikii [3]3 years ago

5 0

Let s = rate of rotation
<span>Let r = radius of earth = 6,400km </span>
<span>Then solving (s^2) r = g will give the desired rate, from which length of day is inferred. </span>
<span>People would not be thrown off. They would simply move eastward in a straight line while the curved surface of earth fell away from beneath them.</span>

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<em></em>

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Answer:

See Explanation

Explanation:

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$s=s_0$

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Solving (b): Units and dimension of $v_0$

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$s=v_0t$

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$v_0t = s$

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$v_0 = \frac{L}{T}$

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Solving (c): Units and dimension of $a_0$

Remove the other terms of the equation, we have:

$s=\frac{a_0t^2}{2}$

Rewrite as:

$\frac{a_0t^2}{2} = s_0$

Make $a_0$ the subject

$a_0 = \frac{2s_0}{t^2}$

Replace s and t with their units [ignore all constants]

$a_0 = \frac{L}{T^2}\\$

$a_0 = LT^{-2$

Hence:

$a_0 = LT^{-2$ --- dimension

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Solving (d): Units and dimension of $j_0$

Remove the other terms of the equation, we have:

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Rewrite as:

$\frac{j_0t^3}{6} = s$

Make $j_0$ the subject

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Replace s and t with their units [Ignore all constants]

$j_0 = \frac{L}{T^3}$

$j_0 = LT^{-3}$

Hence:

$j_0 = LT^{-3}$ --- dimension

$j_0 \to$ $m/s^3$ --- unit

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Remove the other terms of the equation, we have:

$s=\frac{S_0t^4}{24}$

Rewrite as:

$\frac{S_0t^4}{24} = s$

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$S_0 = \frac{L}{T^4}$

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