Answer:
3.64×10⁸ m
3.34×10⁻³ m/s²
Explanation:
Let's define some variables:
M₁ = mass of the Earth
r₁ = r = distance from the Earth's center
M₂ = mass of the moon
r₂ = d − r = distance from the moon's center
d = distance between the Earth and the moon
When the gravitational fields become equal:
GM₁m / r₁² = GM₂m / r₂²
M₁ / r₁² = M₂ / r₂²
M₁ / r² = M₂ / (d − r)²
M₁ / r² = M₂ / (d² − 2dr + r²)
M₁ (d² − 2dr + r²) = M₂ r²
M₁d² − 2dM₁ r + M₁ r² = M₂ r²
M₁d² − 2dM₁ r + (M₁ − M₂) r² = 0
d² − 2d r + (1 − M₂/M₁) r² = 0
Solving with quadratic formula:
r = [ 2d ± √(4d² − 4 (1 − M₂/M₁) d²) ] / 2 (1 − M₂/M₁)
r = [ 2d ± 2d√(1 − (1 − M₂/M₁)) ] / 2 (1 − M₂/M₁)
r = [ 2d ± 2d√(1 − 1 + M₂/M₁) ] / 2 (1 − M₂/M₁)
r = [ 2d ± 2d√(M₂/M₁) ] / 2 (1 − M₂/M₁)
When we plug in the values, we get:
r = 3.64×10⁸ m
If the moon wasn't there, the acceleration due to Earth's gravity would be:
g = GM / r²
g = (6.672×10⁻¹¹ N m²/kg²) (5.98×10²⁴ kg) / (3.64×10⁸ m)²
g = 3.34×10⁻³ m/s²
W = F * d
W = 10N * 2.5 m
W = 25 N m
So the answer you want is the third one down.
Consider a car<span> that travels between points A and B. The </span>car's<span> average </span>speed<span> can be ..... the </span>car<span> to </span>slow down<span> with a </span>constant acceleration<span> of </span>magnitude 3.50 m/s2<span>. </span>If<span> the </span>car comes<span> to a </span>stop<span> in a </span>distance<span> of</span>30.0 m<span>, what was the </span>car's original speed<span>? ... A </span>car<span> is </span>traveling<span> at 26.0 </span>m<span>/s when the </span>driver suddenly applies<span> the </span>brakes<span>, ...</span>
The formula for speed is s = d/t, where s is speed, d is distance, and t is time. The formula can be applied to all objects, including cars, to find their speed.
