Answer:
Approximately 
.
Assumption: the ball dropped with no initial velocity, and that the air resistance on this ball is negligible.
Explanation:
Assume the air resistance on the ball is negligible. Because of gravity, the ball should accelerate downwards at a constant 
near the surface of the earth.
For an object that is accelerating constantly,
,
where
is the initial velocity of the object,
is the final velocity of the object.
is its acceleration, and
is its displacement.
In this case, is the same as the change in the ball's height: 
. By assumption, this ball was dropped with no initial velocity. As a result, 
. Since the ball is accelerating due to gravity, 
.
.
In this case, would be the velocity of the ball just before it hits the ground. Solve for
.
.
Answer:
0.301 m
Explanation:
Torque = Force × Radius
τ = Fr
40.0 Nm = 133 N × r
r = 0.301 m
The mechanic must apply the force 0.301 m from the nut.
Answer:
Ф_cube /Ф_sphere = 3 /π
Explanation:
The electrical flow is
Ф = E A
where E is the electric field and A is the surface area
Let's shut down the electric field with Gauss's law
Фi = ∫ E .dA = 
/ ε₀
the Gaussian surface is a sphere so its area is
A = 4 π r²
the charge inside is
q_{int} = Q
we substitute
E 4π r² = Q /ε₀
E = 1 / 4πε₀ Q / r²
To calculate the flow on the two surfaces
* Sphere
Ф = E A
Ф = 1 / 4πε₀ Q / r² (4π r²)
Ф_sphere = Q /ε₀
* Cube
Let's find the side value of the cube inscribed inside the sphere.
In this case the radius of the sphere is half the diagonal of the cube
r = d / 2
We look for the diagonal with the Pythagorean theorem
d² = L² + L² = 2 L²
d = √2 L
we substitute
r = √2 / 2 L
r = L / √2
L = √2 r
now we can calculate the area of the cube that has 6 faces
A = 6 L²
A = 6 (√2 r)²
A = 12 r²
the flow is
Ф = E A
Ф = 1 / 4πε₀ Q/r² (12r²)
Ф_cubo = 3 /πε₀ Q
the relationship of these two flows is
Ф_cube /Ф_sphere = 3 /π
It is the effect of the Big Bang and dark energy.
V=IR, therefore when resistance is constant the voltage and current are directly proportional
