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
Answer:
Explanation:
Since the equation for the illumination of an object, i.e. the brightness of the light, is <em>inversely proportional to the square of the distance from the light source</em>, the form of the function is:
Where x is the distance between the object and the light force, k is the constant of proportionality, and f(x) is the brightness.
Then, if you move halfway to the lamp the new distance is x/2 and the new brightness (call if F) is :
Then, you have found that the light is 4 times as bright as it originally was.
Answer:
If earth had no tilt, we would have no seasons.
Explanation:
As stated in the answer, if the earth had no tilt we wouldn't have seasons. The earth all around the globe would maintain the same temperature,
And due to the no tilt it would also change our orbit to a bit larger slant, in January when we are at our closest to the sun we WOULD have a mini summer. For the North and South Pole, they would remain cold.
Answer:
V is approximately = 23m/s
Explanation:
Kinetic energy = ½ mv²
Where m= mass = 0.450kg
V= velocity =?
K. E = 119J
Therefore
K. E = ½ mv²
Input values given
119= ½ × 0.450 × v²
Multiply both sides by 2
119 ×2 = 2 × 1/2 × 0.450 × v²
238= 0.450v²
Divide both sides by 0.450
238/0.450 = 0.450v²/0.450
v² = 528.89
Square root both sides
Sq rt v² = sq rt 528.89
V = 22.998m/s
V is approximately = 23m/s
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