Passengers in an aircraft are subject to the Normal and Gravity Force acting on them at a low 'orbit', so tiny that it can be many times compared to the same surface of the earth when speaking in general terms.
In a high orbit space vehicle or in the same space, said force decreases considerably or simply disappears, generating the sensation of weightlessness.
Remember that the Force of Gravity is given under the principle
Where,
G = Gravitational Universal constant
M = Mass of the planet
m = mass of the object
r = Distance from center of the planet
When the radius grows considerably the gravitational force begins to decrease.
Answer:
0.6 m
Explanation:
When a spring is compressed it stores potential energy. This energy is:
Ep = 1/2 * k * x^2
Being x the distance it compressed/stretched.
When the spring bounces the ice cube back it will transfer that energy to the cube, it will raise up the slope, reaching a high point where it will have a speed of zero and a potential energy equal to what the spring gave it.
The potential energy of the ice cube is:
Ep = m * g * h
This is vertical height and is related to the distance up the slope by:
sin(a) = h/d
h = sin(a) * d
Replacing:
Ep = m * g * sin(a) * d
Equating both potential energies:
1/2 * k * x^2 = m * g * sin(a) * d
d = (1/2 * k * x^2) / (m * g * sin(a))
d= (1/2 * 25 * 0.1^2) / (0.05 * 9.81 * sin(25)) = 0.6 m
Answer:
R₂ / R₁ = D / L
Explanation:
The resistance of a metal is
R = ρ L / A
Where ρ is the resistivity of aluminum, L is the length of the resistance and A its cross section
We apply this formal to both configurations
Small face measurements (W W)
The length is
L = W
Area
A = W W = W²
R₁ = ρ W / W² = ρ / W
Large face measurements (D L)
Length L = D= 2W
Area A = W L
R₂ = ρ D / WL = ρ 2W / W L = 2 ρ/L
The relationship is
R₂ / R₁ = 2W²/L
Answer:
0.5*10uF * 16*16 =0.0128
Explanation:
I have no explanation just like my soul.
