the answer is c) the speed and direction of travel must be constant
Answer:
Explanation:
Let electric potential at A ,B and C be Va , Vb and Vc respectively.
Work done = charge x potential difference
Wab = q ( Va - Vb )
Wac = q ( Va - Vc )
Given
Wac = - Wab / 3
3Wac = - Wab
Now
Wbc = q ( Vb - Vc )
= q [ ( Va-Vc ) - ( Va - Vb )]
= Wac - Wab
= Wac + 3Wac
= 4Wac
Since my givens are x = .550m [Vsub0] = unknown
[Asubx] = =9.80
[Vsubx]^2 = [Vsub0x]^2 + 2[Asubx] * (X-[Xsub0]
[Vsubx]^2 = [Vsub0x]^2 + 2[Asubx] * (X-[Xsub0])
Vsubx is the final velocity, which at the max height is 0, and Xsub0 is just 0 as that's where it starts so I just plug the rest in
0^2 = [Vsub0x]^2 + 2[-9.80]*(.550)
0 = [Vsub0x]^2 -10.78
10.78 = [Vsub0x]^2
Sqrt(10.78) = 3.28 m/s
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