I think the situation is modeled by the scenario in the attached image. Some specific values seem to be missing (like the height of door 
)...
The door forms a right triangles that satisfies
We also have
so if you happen to know the height of the door, you can solve for 
and 
.
is fixed, so
We can solve for the angular velocity 
:
At the point when 
and 
ft/s, we get
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:
150156.25 Ω
Explanation:
Resistance: This can be defined as the opposition to the flow of electric current in a circuit. The S.I unit of resistance is Ohm's (Ω)
The expression for resistance is given as
P = V²/R................ equation 1
Where P = power, V = Voltage, R = Resistance.
Making R the subject of the equation,
R = V²/P.................. Equation 2
Given: V = 115 V , P = 0.16 W.
Substitute into equation 2
R = 155²/0.16
R = 150156.25 Ω
Hence,
The resistance = 150156.25 Ω
