Clarify what you mean by ratios?
Answer:
a) The current density ,J = 2.05×10^-5
b) The drift velocity Vd= 1.51×10^-15
Explanation:
The equation for the current density and drift velocity is given by:
J = i/A = (ne)×Vd
Where i= current
A = Are
Vd = drift velocity
e = charge ,q= 1.602 ×10^-19C
n = volume
Given: i = 5.8×10^-10A
Raduis,r = 3mm= 3.0×10^-3m
n = 8.49×10^28m^3
a) Current density, J =( 5.8×10^-10)/[3.142(3.0×10^-3)^2]
J = (5.8×10^-10) /(2.83×10^-5)
J = 2.05 ×10^-5
b) Drift velocity, Vd = J/ (ne)
Vd = (2.05×10^-5)/ (8.49×10^28)(1.602×10^-19)
Vd = (2.05×10^-5)/(1.36 ×10^10)
Vd = 1.51× 10^-5
Answer:
1069.38 gallons
Explanation:
Let V₀ = 1.07 × 10³ be the initial volume of the gasoline at temperature θ₁ = 52 °F. Let V₁ be the volume at θ₂ = 97 °F.
V₁ = V₀(1 + βΔθ) β = coefficient of volume expansion for gasoline = 9.6 × 10⁻⁴ °C⁻¹
Δθ = (5/9)(97°F -52°F) °C = 25 °C.
Let V₂ be its final volume when it cools to 52°F in the tank is
V₂ = V₁(1 - βΔθ) = V₀(1 + βΔθ)(1 - βΔθ) = V₀(1 - [βΔθ]²)
= 1.07 × 10³(1 - [9.6 × 10⁻⁴ °C⁻¹ × 25 °C]²)
= 1.07 × 10³(1 - [0.024]²)
= 1.07 × 10³(1 - 0.000576)
= 1.07 × 10³(0.999424)
= 1069.38 gallons
Answer:
6.0 × 
W/
Explanation:
From Wien's displacement formula;
Q = e A
Where: Q is the quantity of heat transferred, e is the emissivity of the surface, A is the area, and T is the temperature.
The emissive intensity = 
= e
Given from the question that: e = 0.6 and T = 1000K, thus;
emissive intensity = 0.6 × 
= 0.6 × 1.0 × 
= 6.0 × 
Therefore, the emissive intensity coming out of the surface is 6.0 × W/.
Answer:
Capacitor
Explanation:
Capacitor ; A capacitor is a passive two-terminal electrical component that can store energy in an electric field electrostatically. It works as a small rechargeable battery that stores electricity. However, unlike a battery, it can charge and discharge in the split of a second. Capacitors are widely used to build different types of electronic circuits.
