Which of the following statements about price elasticity of demand is true?
1 answer:
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Answer:
The maximum induced emf in the rotating coil = 29.66V
The induced emf in the rotating coil when (t = 1.00 s) = 26.66V
The maximum rate of change of the magnetic flux through the rotating coil = 0.674Wb/s
Explanation:
Lets state the parameters we are being given right from the question:
Number of rectangular coil, (N) = 44
Length of Coil, l =17cm in meters we have; (l) = 17 × 10⁻² m
Width of Coil, w =8.10cm in meters we have; (w) = 8.10 × 10⁻² m
Magnitude of Uniform Magnetic Field (B) = 767mT= 765 × 10⁻³ T
Angular Speed of Coil, (ω) = 64 rad/s
(a)
To calculate the induced emf in the rotating cell,we can use the formula:
emf = NBAωsin(ωt)
For maximum induced emf, the value of sin(ωt) will be 1
= NBAω ; if (A = l × w) , we have:
= NB(l × w)ω
subsitituting the parameters into the above equation; we have:
= 44 × 765 × 10⁻³ ( 17 × 10⁻² × 8.10 × 10⁻² ) × 64
= 29.66V
(b)
At t = 1s, the induced emf is calculated as:
emf = NBAωsin(ωt)
substituting the parameters into the equation, we have:
emf = 44 × 765 × 10⁻³ ( 17 × 10⁻² × 8.10 × 10⁻² ) × 64 × sin (64 × 1)
=26.66V
(c)
To calculate the maximum rate of change of the magnetic flux through the rotating coil; we need to reflect on the equation for the maximum induced emf in terms of magnetic flux.
i.e =
since = 29.66 and N = 44; we have:
29.66 =
=
= 0.674 Wb/s
Answer:
a. The volume V₁ and V₂
b. The case that involves the greatest momentum change = Case B
c. The case that involves the greatest impulse = Case B
d. b. The case that involves the greatest force = Case B
Explanation:
Here we have
Case A: V₁ = 150-mL, v₁ = 8 m/s
Case B: V₂ = 600-mL, v₁ = 8 m/s
a. The variable that is different for the two cases is the volume V₁ and V₂
b. The momentum change is by the following relation;
ΔM₁ = Mass, m × Δv₁
The mass of the balloon are;
Δv₁ = Change in velocity = Final velocity - Initial velocity
Mass = Density × Volume
Density of water = 0.997 g/mL
Case A, mass = 150 × 0.997 = 149.55 g
Case B, mass = 600 × 0.997 = 598.2 g
The momentum change is;
Case A: Mass, m × Δv₁ = 149.55 g/1000 × 8 m/s = 1.1964 g·m/s
Case B: Mass, m × Δv₁ = 598.2/1000 × 8 = 4.7856 g·m/s
Therefore Case B has the greatest momentum change
The case that has the gretest momentum change = Case B
c. The momentum change = impulse therefore Case B involves the greatest impulse
d. Here we have;
Impulse = Momentum change = × Δt = mΔV
∴ = m·ΔV/Δt
∴ For Case A = 149.55×8/Δt = 1196.4/Δt N
For Case B = 598.2×8/Δt = 4785.6/Δt
Where Δt is the same for Case A and Case B, for Case B >>
for Case B
Therefore, Case B involves the greatest force.