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
<span>The potential energy decreases liquid particles that are attracted to one another move closer together to form a solid. Temperature remains constant because kinetic energy remains constant</span>
Answer:
v₁ = 37.5 cm / s
Explanation:
For this exercise we can use that angular and linear velocity are related
v = w r
in the case of the spool the angular velocity for the whole system is constant,
They indicate the linear velocity v₀ = 25.0 cm / s for a radius of r₀ = 1.00 cm,
w = v₀ /r₀
for the outside of the spool r₁ = 1.5 cm
w = v₁ / r₁1
since the angular velocity is the same we set the two expressions equal
v1 = 
let's calculate
v₁ = 
v₁ = 37.5 cm / s
The coordinates are- <span>RA 19° 24' 0" | Dec 59° 0.000'
It is located in the asteroid belt between mars and jupiter :)
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1 hour = 3600 seconds.
Energy dissipated = I²Rt = 8²×20×3600 = 4608000 J
