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2 years ago

A man weighs 90 N on the surface of earth of radius R at what height above the surface of earth is weight will be 40 N.

Physics

1 answer:

lapo4ka [179]2 years ago

4 0

Answer:

3/2 R

Explanation:

Let the Mass of man be m, Mass of earth be M.

By Newtons, law of gravitation,

F = GMm/R^2, Where G is gravitational constant

90 = GMm/R^2

When man weighs 40N,

GMm/(New R)^2 = 40

= 4/9 (90)

GMm/(New R)^2 = 4/9 (GMm/R^2)

(New R)^2 = 9/4 R^2

New R = 3/2 R

Therefore, height is 3/2 R

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liq [111]

Answer:

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2 years ago

*please refer to photo* An electric field of magnitude 5.25 ✕ 10^5N/C points due south at a certain location. Find the magnitude

kvv77 [185]

Answer:

Approximately $3.86\; {\rm N}$ (given that the magnitude of this charge is $-7.35\; {\rm \mu C}$ .)

Explanation:

If a charge of magnitude $q$ is placed in an electric field of magnitude $E$ , the magnitude of the electrostatic force on that charge would be $F = E\, q$ .

The magnitude of this charge is $q = 7.35\; {\rm \mu C}$ . Apply the unit conversion $1\; {\rm \mu C} = 10^{-6}\; {\rm C}$ :

$\begin{aligned} q &= 7.35\; {\mu C} \times \frac{10^{-6}\; {\rm C}}{1\; {\mu C}} = 7.35\times 10^{-6}\; {\rm C}\end{aligned}$ .

An electric field of magnitude $E = 5.25\times 10^{5}\; {\rm N \cdot C^{-1}}$ would exert on this charge a force with a magnitude of:

$\begin{aligned}F &= E\, q \\ &= 5.25 \times 10^{5}\; {\rm N \cdot C^{-1}} \times (-7.35\times 10^{-6}\; {\rm C}) \\ &\approx 3.86\; {\rm N}\end{aligned}$ .

Note that the electric charge in this question is negative. Hence, electrostatic force on this charge would be opposite in direction to the the electric field. Since the electric field points due south, the electrostatic force on this charge would point due north.

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1 year ago

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Nikolay [14]

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2 years ago

If a stone is dropped from a height of 400 feet, its height after t seconds is given by s = 400 − 16t2. Find its instantaneous v

soldier1979 [14.2K]

Answer:

$v = -32 t$