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3 years ago

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By how many newtons does the weight of a 85.9-kg person lose when he goes from sea level to an altitude of 6.33 km if we neglect

the earth's rotational effects

1 answer:

sergejj [24]3 years ago

3 0

Answer:

$Weight\ loss=1.6321N$

Explanation:

From the question we are told that:

Weight $W=85.9kg$

Altitude $h= 6.33 km$

Let

Radius of Earth $r=6380km$

Gravity $g=9.8m/s^2$

Generally the equation for Gravity at altitude is mathematically given by

$g_s=9.8(\frac{6380}{6380+6.33})^2$

$g_s=9.781m/s^2$

Therefore

Weight at sea level

$W_s=9.8*85.9$

$W_s=841.82N$

Weight at 6.33 altitude

$W_a=9.781*85.9$

$W_a=840.2N$

Therefore

$Weight loss=W_s-W_b$

$Weight loss=841.82-840.2$

$Weight loss=1.6321N$

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Explanation:

Given that,

(a) Work done by the electric field is 12 J on a 0.0001 C of charge. The electric potential is defined as the work done per unit charged particles. It is given by :

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(b) Similarly, same electric field does 24 J of work on a 0.0002-C charge. The electric potential difference is given by :

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Answer:

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2 years ago

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Answer:

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Explanation:

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Answer:

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3 years ago

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