Power is force multiplied by velocity. The engine power is actually (relatively) constant regardless of the gear. So when people say there is "more power" in a lower gear, it's the common misconception that "more powerful" is "more forceful" but that's only part of the equation.
So if P is constant, then that means if you can combine a large force and a low velocity or a low force and large velocity for the same power.
When you put it in a low gear, you produce a large torque -- or a large force -- and a low velocity.
For example, if you are towing a trailer or trying to climb a very steep grade, you need the force to be large which is why you put it in a low gear. If you are on something slippery like snow or ice, a high gear will keep the force at the wheels low so the tires don't exceed the coefficient of friction and spin.
The numbers of records that will be displayed in response to his query is known to be zero (0).
<h3>What is a query?</h3>
A query is known to be a kind of a question or any form of request that is known to be for information that is often expressed in a formal manner.
Note that a database query is one that can be seen as an action query or a a kind of select query.
A select query is one that can help to get back data from a database. based on the question, Jeremy is looking for a brown Tennessee walker and this is not among the option presented. Therefore the query will come back as zero.
Hence, The numbers of records that will be displayed in response to his query is known to be zero (0).
Learn more about query from
brainly.com/question/25694408
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the answer in my opinion would have to be B.
Answer:
Here it is
Explanation:
#include <iostream>
using namespace std;
int main()
{
int a, b;
cin >> a >> b;
int nr_digits1 = 0, nr_digits2 = 0;
int sum1 = 0, sum2 = 0;
int max_digit1 = 0, max_digit2 = 0;
while (a > 1)
{
int digit = a % 10;
nr_digits1 += 1;
sum1 += digit;
if (digit > max_digit1)
{
max_digit1 = digit;
}
a /= 10;
}
while (b > 1)
{
int digit = b % 10;
nr_digits2 += 1;
sum2 += digit;
if (digit > max_digit2)
{
max_digit2 = digit;
}
b /= 10;
}
cout << "For a: \n" << " No. of digits: " << nr_digits1 << "\n";
cout << " Sum of digits: " << sum1 << "\n";
cout << " Max digit: " << max_digit1 << "\n";
cout << "\n";
cout << "For b: \n" << " No. of digits: " << nr_digits2 << "\n";
cout << " Sum of digits: " << sum2 << "\n";
cout << " Max digit: " << max_digit2 << "\n";
return 0;
}
Valid
$4 , apps , miles , x
keywrd:
class , int
