Answer:
The voltage needed to accelerate the electron beam is 2.46 x 10^16 Volts
Explanation:
The rate of electron flow is given as:
q = 1015 electrons per second
The total current is given by:
Total Current = (Rate of electron flow)(Charge on one electron)
Total Current = I = (1015 electrons/s)(1.6 x 10^-19 C/electron)
I = 1.624 x 10^-16 A
Now, we know that electric power is given as:
Electric Power = Current x Voltage
P = IV
V = P/I
V = 4 W/1.624 X 10^-16 A
<u>V = 2.46 x 10^16 Volts</u>
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:
screw is the answer of the question
Answer:
def output_ints_less_than_or_equal_to_threshold(user_values, upper_threshold):
for value in user_values:
if value < upper_threshold:
print(value)
def get_user_values():
n = int(input())
lst = []
for i in range(n):
lst.append(int(input()))
return lst
if __name__ == '__main__':
userValues = get_user_values()
upperThreshold = int(input())
output_ints_less_than_or_equal_to_threshold(userValues, upperThreshold)