The magnetic field or force seems to be associated with the lineup of electrons withim the magnet
The energy transfer in terms of work has the equation:
W = mΔ(PV)
To be consistent with units, let's convert them first as follows:
P₁ = 80 lbf/in² * (1 ft/12 in)² = 5/9 lbf/ft²
P₂ = 20 lbf/in² * (1 ft/12 in)² = 5/36 lbf/ft²
V₁ = 4 ft³/lbm
V₂ = 11 ft³/lbm
W = m(P₂V₂ - P₁V₁)
W = (14.5 lbm)[(5/36 lbf/ft²)(4 ft³/lbm) - (5/9 lbf/ft²)(11 lbm/ft³)]
W = -80.556 ft·lbf
In 1 Btu, there is 779 ft·lbf. Thus, work in Btu is:
W = -80.556 ft·lbf(1 Btu/779 ft·lbf)
<em>W = -0.1034 BTU</em>
Answer:
a. λ = 647.2 nm
b. I₀ 9.36 x 10⁻⁵
Explanation:
Given:
β = 56.0 rad , θ = 3.09 ° , γ = 0.170 mm = 0.170 x 10⁻³ m
a.
The wavelength of the radiation can be find using
β = 2 π / γ * sin θ
λ = [ 2π * γ * sin θ ] / β
λ = [ 2π * 0.107 x 10⁻³m * sin (3.09°) ] / 56.0 rad
λ = 647.14 x 10⁻⁹ m ⇒ λ = 647.2 nm
b.
The intensity of the central maximum I₀
I = I₀ (4 / β² ) * sin ( β / 2)²
I = I₀ (4 / 56.0²) * [ sin (56.0 /2) ]²
I = I₀ 9.36 x 10⁻⁵
Answer:
I = 0.2 A
Explanation:
Lamp is rated at 300 mA
I_lamp = 0.3 A
Voltage is; V = 3V
Thus; Resistance is given by;
R = V/I
R = 3/0.3
R = 10 ohms
Now, since the ammeter of 5 ohms is connected in series with the lamp. Thus equivalent resistance;
R_eq = 10 + 5
R_eq = 15 ohms
Ammeter current will be;
I = V/R_eq
I = 3/15
I = 0.2 A
