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3 years ago

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A 5.0-kg rock and a 3.0 × 10−4-kg pebble are held near the surface of the earth.(a)Determine the magnitude of the gravitational

force exerted on each by the earth.(b)Calculate the magnitude of the acceleration of each object when released.

2 answers:

liraira [26]3 years ago

4 0

Answer:

A) Gravitational Force of Rock = 48.995N

Gravitational Force of Pebble = 2.94 x 10^(-3) N

B) Magnitude of acceleration of rock = 9.8 m/s²

Magnitude of acceleration of pebble = 9.8 m/s²

Explanation:

A) First of all, let's solve for the rock;

From Newton's law of universal gravitation, gravitational force is;

Fg = (Gm1m2)/r²

Where G is a constant = 6.67 × 10^(-11) Nm² /kg2²

Mass of earth (m1) = 5.97 × 10^24 kg

Radius of earth (r) = 6.38 × 10^6 m²

Mass of rock (m2) = 5kg.

Thus, Fg = [6.67 × 10^(-11) x 5.97 × 10^24 x 5] / (6.38 × 10^6)² = 48.995N

Doing the same for the pebble of mass 3 x 10^(-4) kg;

Thus, Fg = [6.67 × 10^(-11) x 5.97 × 10^(24) x 3 x 10^(-4)] / (6.38 × 10^6)² = 2.94 x 10^(-3) N

B) Let's calculate for the acceleration of the rock;

We know that Fg = ma

Thus, Let's make a the subject of the formula,

So, Fg/m = a

Substituting rock value of Fg and m to get;

48.995/5 = 9.8 m/s²

Now let's do the same for the pebble;

Fg/m = a

Substituting pebble value of Fg and m to get;

[2.94 x 10^(-3) N] / (3 x 10^(-4)) = 9.8 m/s²

a_sh-v [17]3 years ago

3 0

Answer:

a). Determine the magnitude of the gravitational force exerted on each by the earth.

Rock: $F = 49.06N$

Pebble: $F = 29.44N$

(b)Calculate the magnitude of the acceleration of each object when released.

Rock: $a =9.8m/s^{2}$

Pebble: $a =9.8m/s^{2}$

Explanation:

The universal law of gravitation is defined as:

$F = G\frac{m1m2}{r^{2}}$ (1)

Where G is the gravitational constant, m1 and m2 are the masses of the two objects and r is the distance between them.

<em>Case for the rock </em> $m = 5.0 Kg$ <em>:</em>

m1 will be equal to the mass of the Earth $m1 = 5.972×10^{24} Kg$ and since the rock and the pebble are held near the surface of the Earth, then, r will be equal to the radius of the Earth $r = 6371000m$ .

$F = (6.67x10^{-11}kg.m/s^{2}.m^{2}/kg^{2})\frac{(5.972x10^{24} Kg)(5.0 Kg)}{(6371000 m)^{2}}$

$F = 49.06N$

Newton's second law can be used to know the acceleration.

$F = ma$

$a =\frac{F}{m}$ (2)

$a =\frac{(49.06 Kg.m/s^{2})}{(5.0 Kg)}$

$a =9.8m/s^{2}$

<em>Case for the pebble </em> $m = 3.0 Kg$ <em>:</em>

$F = (6.67x10^{-11}kg.m/s^{2}.m^{2}/kg^{2})\frac{(5.972x10^{24} Kg)(3.0 Kg)}{(6371000 m)^{2}}$

$F = 29.44N$

$a =\frac{F}{m}$

$a =\frac{(29.44 Kg.m/s^{2})}{(3.0 Kg)}$

$a =9.8m/s^{2}$

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