Answer:A)Engine flywheel
Explanation: Clutch is the part of the vehicles like car,motorbike etc that persist the property of the engaging and non-engaging of the engine from the wheels.Manual clutch is handled by the driver of the vehicle.
The clutch is embedded on the flywheel which is a heavy device connected with engine. This part stores some amount of energy and works on the delivery of the power.Thus the correct option is option (a) as the other options are not the defined parts of the vehicle.
Answer:
the correct distance is 202 ft
Explanation:
The computation of the correct distance is shown below:
But before that correction to be applied should be determined
= (101 ft - 100 ft) ÷ (100 ft) × 200 ft
= 2 ft
Now the correct distance is
= 200 ft + 2 ft
= 202 ft
Hence, the correct distance is 202 ft
The same would be relevant and considered too
Answer:
The program is given below with appropriate comments for better understanding
Explanation:
#Program
# foot stride = 2.5 feet
# 1 mile = 5280 feet
no_stride_first_min = int(input('Enter the number strides made durng the first minute of jogging: '))
no_stride_last_min = int(input('Enter the number strides made durng the last minute of jogging: '))
avg_stride_one_min = (no_stride_first_min + no_stride_last_min)/2 # calculates the average stride per minute
jogging_duration = float(input('Enter the total time spent jogging in hours and minute: '))
jogging_duration_hours = int(jogging_duration) # gets the hour
jogging_duration_min = jogging_duration - int(jogging_duration) # gets the minute
tot_jogging_duration_min = jogging_duration_hours*60 + jogging_duration_min # calculates total time in minutes
dist_feet = (avg_stride_one_min*2.5)*tot_jogging_duration_min # calculates the total distance in feet
dist_miles = dist_feet/5280 # calculates the total distance in mile
print('Distance traveled in miles = {0:.2f} miles'.format(dist_miles))
Answer:
A. Manufacturers rating capacity ↔ 3. Must be marked on all jacks; must not be exceeded
B. Block Used to lift and hold heavy loads, allow them for travel ↔ 1. Place the jack head against this
C. Level surface ↔ 4. Place this under the base of the jack when it's necessary to provide a firm foundation
D. Jack ↔ 2. Used to lift and hold heavy loads, allow them for travel
Explanation:
The manufacturers rating for a jack is labelled on all jacks and should be referenced to compare with the load to be lifted so as to ensure a safe and successful lifting.
In order to lift a load, such as a car, it is required to place the jack on a level surface to provide balance during the lifting task
The head of the jack is placed against the block for lifting heavy objects for proper performance
Answer:
channel width = 2.621 ft
Explanation:
Given data :
Decreasing Factor = 2
subcritical value = 0.4
super critical value = 2.5
width = 13 ft
attached below is a detailed solution and