Answer:
676 ft
Explanation:
Minimum sight distance, d_min
d_min = 1.47 * v_max * t_total where v_max is maximum velocity in mi/h, t_total is total time
v_max is given as 50 mi/h
t_total is sum of time for right-turn and adjustment time=8.5+0.7=9.2 seconds
Substituting these figures we obtain d_min=1.47*50*9.2=676.2 ft
For practical purposes, this distance is taken as 676 ft
Answer:
1) 63.66 ohm
2) 188.49 ohm
Explanation:
Data provided in the question:
Part 1
Capacitance, C = 5μF = 5 × 10⁻⁶ F
Frequency = 500 Hz
Now,
Impedance = 
or
Impedance = 
or
Impedance = 63.66 ohm
Part 2
Inductance = 60 mH = 60 × 10⁻³ H
Frequency = 500 Hz
Now,
Impedance for an inductor = 2πfL
thus,
Impedance = 2 × π × 500 × 60 × 10⁻³
= 188.49 ohm
Explanation:
Okay soo-
Given-
u = 60 km/hr = 60×1000/3600=50/3 m/s
t = 20 s
s = 250 m
a = ?
v = ?
Solution -
Here, acceleration is uniform.
(a) According to 2nd kinematics equation,
s = ut + ½at^2
250 = 50/3 ×20 + 0.5×a×20×20
250-1000/3=200a
(750-1000)/3=200a
a = -250/(3×200)
a = -5/12
a = 0.4167 m/s^2
The required uniform acceleration of the car is 0.4167 m/s^2.
(b) According to 1st kinematics equation
v = u + at
v = 50/3 + (-5/12)×20
v = 50/3-25/3
v = 25/3
v = 8.33 m/s
The speed of the car as it passes the traffic light is 8.33 m/s.
Good luck!
