A car travels a distance of 140 km at 70.0 km/hr. It then travels an additional distance of 60.0 km at 40.0 km/hr. The average s
1 answer:
You might be interested in
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
Answer:
T1 = 131.4 [N]
T2 = 261 [N]
Explanation:
To solve this problem we must make a sketch of how will be the semicircle, for this reason we conducted an internet search, to find the scheme of the problem. This scheme is attached in the first image.
Then we make a free body diagram, with this free body diagram, we raise the forces that act on the body. Since it is a problem involving static equilibrium, the sum of forces in any direction and moments must be equal to zero.
By performing a sum of forces on the Y axis equal to zero we can find an equation that relates the forces of tension T1 & T2.
The second equation can be determined by summing moments equal to zero, around the point of application of the T1 force. In this way we find the T2 force.
The value of T2, is replaced in the first equation and we can find the value for T1.
Therefore
T1 = 131.4 [N]
T2 = 261 [N]
The free body diagram and the developed equations can be seen in the second attached image.
