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How many non-square numbers lie between the squares of 12 and 13?
Answer
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Hint: Here, we can see that 12 and 13 are consecutive numbers. So, all numbers between squares of 12 and 13 are non-square numbers. Therefore, first find squares of 12 and 13 and then subtract square of 12 from square of 13, we get numbers of non-square numbers. At the last subtract 1 from the result obtained as both extremes numbers are not included.
Complete step-by-step answer:
In these types of questions, a simple concept of numbers should be known that is between squares of two consecutive numbers all numbers are non-square numbers. Also one tricky point should remember that whenever we find the difference between two numbers we get a number of numbers between them including anyone of the extreme numbers. So we subtract 1 to exclude both extreme numbers.
Square of 12 = 122=144 and square of 12 = 132=169
As 12 and 13 are consecutive numbers so all numbers between their squares will be non-square numbers.
Therefore, 169 – 144 = 25
Total number of numbers between 169 and 144 (i.e., excluding 144 and 169) = 25 – 1 = 24.
Explanation:
Brian least po please
C because the the smallest thing then the other ones because it never said what kind of size of it
Explanation:
It varies with altitude, but at sea level, it's 9.8 m/s².
The cart will be pulled to the right by the hanging mass, so by Newton's second law, the net force on the cart is
<em>T</em> - 25 N = (8 kg) <em>a</em>
where <em>T</em> is the tension in the rope and <em>a</em> is the acceleration.
The hanging mass has a net force of
(6 kg) <em>g</em> - <em>T</em> = (6 kg) <em>a</em>
where <em>g</em> = 9.8 m/s².
Adding these equations together eliminates <em>T</em>, and we can solve for <em>a</em> :
(<em>T</em> - 25 N) + ((6 kg) <em>g</em> - <em>T </em>) = (14 kg) <em>a</em>
33.8 N = (14 kg) <em>a</em>
<em>a</em> = (33.8 N) / (14 kg) ≈ 2.4 m/s²
Then the tension in the rope is
<em>T</em> - 25 N = (8 kg) (2.4 m/s²)
<em>T</em> ≈ 25 N + 19.31 N ≈ 44 N
