Answer
given,
6 lanes divided highway 3 lanes in each direction
rolling terrain
lane width = 10'
shoulder on right = 5'
PHF = 0.9
shoulder on the left direction = 3'
peak hour volume = 3500 veh/hr
large truck = 7 %
tractor trailer = 3 %
speed = 55 mi/h
LOS is determined based on V p
10' lane weight ; f_{Lw}=6.6 mi/h
5' on right ; f_{Lc} = 0.4 mi/hr
3' on left ; no adjustment
3 lanes in each direction f n = 3 mi/h
= 0.877
= 1,555 veh/hr/lane
= (55 + 5) - 6.6 - 0.4 -3 -0
= 50 mi/h
level of service is D using speed flow curves and LOS for basic free moving of vehicle
Answer:
The energy, that is dissipated in the resistor during this time interval is 153.6 mJ
Explanation:
Given;
number of turns, N = 179
radius of the circular coil, r = 3.95 cm = 0.0395 m
resistance, R = 10.1 Ω
time, t = 0.163 s
magnetic field strength, B = 0.573 T
Induced emf is given as;
where;
ΔФ is change in magnetic flux
ΔФ = BA = B x πr²
ΔФ = 0.573 x π(0.0395)² = 0.002809 T.m²
According to ohm's law;
V = IR
I = V / R
I = 3.0848 / 10.1
I = 0.3054 A
Energy = I²Rt
Energy = (0.3054)² x 10.1 x 0.163
Energy = 0.1536 J
Energy = 153.6 mJ
Therefore, the energy, that is dissipated in the resistor during this time interval is 153.6 mJ
Answer:
A supercapacitor, also called an ultracapacitor, is a high-capacity capacitor with a capacitance value much higher than other capacitors, but with lower voltage limits, that bridges the gap between electrolytic capacitors and rechargeable batteries.
Explanation:
Answer:
2750
Explanation:
The number of windings and the voltage are proportional.
__
Let n represent the number of windings to produce 110 Vac. Then the proportion is ...
n/110 = 300,000/12,000
n = 110(300/12) = 2750 . . . . multiply by 110
2750 windings would be needed to produce 110 Vac at the output.
Answer:
50421.6 m³
Explanation:
The river has an average rate of water flow of 59.6 m³/s.
Tributary B accounts for 47% of the rate of water flow. Therefore the rate of water flow through tributary B is:
Flow rate of water through tributary B = 47% of 59.6 m³/s = 0.47 * 59.6 m³/s = 28.012 m³/s
The volume of water that has been discharged through tributary B = Flow rate of water through tributary B * time taken
time = 30 minutes = 30 minutes * 60 seconds / minute = 1800 seconds
The volume of water that has been discharged through tributary B in 30 seconds = 28.012 m³/s * 1800 seconds = 50421.6 m³
