Driving on under-inflated tires is very dangerous. If tires pressure is too low, too much of the tire's surface area touches the road, which increases friction. The more friction you have, the more heat is applied to the tires. Then, too much heat will make them overheat which can cause tread separation and, yes, blowouts. Therefore, the answer is true.
Answer:
The Session layer (Layer 5)
Explanation:
The OSI (Open Systems Interconnection) seven-layer model is a reference model that explains how devices (or applications) communicate over the network. The model consists basically of seven different layers which are;
i. Layer 1 or the physical layer
ii. Layer 2 or the data-link layer
iii. Layer 3 or the network layer
iv. Layer 4 or the transport layer
v. Layer 5 or the Session layer
vi. Layer 6 or the presentation layer
vii. Layer 7 or the application layer.
Out of the seven layers, the session layer (layer 5) keeps track of a system's connections to send the right response to the right computer by setting up, coordinating and managing these connections. It controls, establishes and maintains the connection between multiple computers or applications on a computer.
Code:
def myAppend( str, ch ):
# Return a new string that is like str but with
# character ch added at the end
return str + ch
def myCount( str, ch ):
# Return the number of times character ch appears
# in str.
# initiaalizing count with 0
count = 0
# iterating over every characters present in str
for character in str:
# incrementing count by 1 if character == ch
if character == ch:
count += 1
# returning count
return count
def myExtend( str1, str2 ):
# Return a new string that contains the elements of
# str1 followed by the elements of str2, in the same
# order they appear in str2.
# concatenating both strings and returning its result
return str1 + str2
def myMin( str ):
# Return the character in str with the lowest ASCII code.
# If str is empty, print "Empty string: no min value"
# and return None.
if str == "":
print("Empty string: no min value")
return None
# storing first character from str in char
char = str[0]
# iterating over every characters present in str
for character in str:
# if current character is lower than char then
# assigning char with current character
if character < char:
char = character
# returning char
return char
def myInsert( str, i, ch ):
# Return a new string like str except that ch has been
# inserted at the ith position. I.e., the string is now
# one character longer than before.
# Print "Invalid index" if
# i is greater than the length of str and return None.
if i > len(str):
print("Invalid index")
return None
# str[:i] gives substring starting from 0 and upto ith position
# str[i:] gives substring starting from i and till last position
# returning the concatenated result of all three
return str[:i]+ch+str[i:]
def myPop( str, i ):
# Return two results:
# 1. a new string that is like str but with the ith
# element removed;
# 2. the value that was removed.
# Print "Invalid index" if i is greater than or
# equal to len(str), and return str unchanged and None
if i >= len(str):
print("Invalid index")
return str, None
# finding new string without ith character
new_str = str[:i] + str[i+1:]
# returning new_str and popped character
return new_str, str[i]
def myFind( str, ch ):
# Return the index of the first (leftmost) occurrence of
# ch in str, if any. Return -1 if ch does not occur in str.
# finding length of the string
length = len(str)
# iterating over every characters present in str
for i in range(length):
# returning position i at which character was found
if str[i]==ch:
return i
# returning -1 otherwise
return -1
def myRFind( str, ch ):
# Return the index of the last (rightmost) occurrence of
# ch in str, if any. Return -1 if ch does not occur in str.
# finding length of the string
length = len(str)
# iterating over every characters present in str from right side
for i in range(length-1, 0, -1):
# returning position i at which character was found
if str[i]==ch:
return i
# returning -1 otherwise
return -1
def myRemove( str, ch ):
# Return a new string with the first occurrence of ch
# removed. If there is none, return str.
# returning str if ch is not present in str
if ch not in str:
return str
# finding position of first occurence of ch in str
pos = 0
for char in str:
# stopping loop if both character matches
if char == ch:
break
# incrementing pos by 1
pos += 1
# returning strig excluding first occurence of ch
return str[:pos] + str[pos+1:]
def myRemoveAll( str, ch ):
# Return a new string with all occurrences of ch.
# removed. If there are none, return str.
# creating an empty string
string = ""
# iterating over each and every character of str
for char in str:
# if char is not matching with ch then adding it to string
if char!=ch:
string += char
# returning string
return string
def myReverse( str ):
# Return a new string like str but with the characters
# in the reverse order.
return str[::-1]
For most usb devices, power is supplied via the port.
