a car enters freeway with speed of 6.4m/s and accelerates uniformly for 3.2 Km in 3.5 min.how fast in m/s is the car moving afte
2 answers:
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Answer:
E = k Q₁ / r²
Explanation:
For this exercise that asks us for the electric field between the sphere and the spherical shell, we can use Gauss's law
Ф = ∫ E .dA = / ε₀
where Ф the electric flow, qint is the charge inside the surface
To solve these problems we must create a Gaussian surface that takes advantage of the symmetry of the problem, in this almost our surface is a sphere of radius r, that this is the sphere of and the shell, bone
R <r <R_a
for this surface the electric field lines are radial and the radius of the sphere are also, therefore the two are parallel, which reduces the scalar product to the algebraic product.
E A = q_{int} /ε₀
The charge inside the surface is Q₁, since the other charge Q₂ is outside the Gaussian surface, therefore it does not contribute to the electric field
q_{int} = Q₁
The surface area is
A = 4π r²
we substitute
E 4π r² = Q₁ /ε₀
E = 1 / 4πε₀ Q₁ / r²
k = 1/4πε₀
E = k Q₁ / r²
Answer:
A= 203 KJ
B= 54 Kg
Explanation:
The initial specific volumes and internal energies are obtained from A-12 for a given pressure and state. The enthalpy of the refrigerant in the supply line is determined using the saturated liquid approximation for the given temperature with data from A-11. The mass that has entered the tank is:
Δm = m₂ – m₁
= V(1/α₂ – 1/α₁)
= 0.05 (1/0.0008935 – 1/ 0.025645)Kg
= 54Kg
The heat transfer is obtained from the energy balance:
ΔU=
+
m₂u₂ – m₁u₂ = +
= m₂u₂ – m₁u₁ –
= V/α₂u₂ - V/α₁u₁ –
=(0.05/0.0008935 . 116.72 – 0.05/0.025645 . 246.82 – 54.108.28) Kj
= 203 KJ