a carmboard 4ft x 4ft Square has the queen at the centre the queen is hit by the striker moves to the front edge, rebounds and g
oes inta hole behind the striking line. find the magnitude of displacement of queen a) from the centre to front edge b) from the front end edge to hole (c) from centre to hole.
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Answer:
Tension T1 is less than tension T2.
T1 < T2
Explanation:
According to given data,
mass of box A ( mA) is grater than mass of box B (mB)
we can write,
m(A) > m(B)
Newton's second law states that:
Tension of object is directly proportional to the mass of the system.
T ∝ m
here Boxes A and B are being pulled to the right on a frictionless surface,
so Tension T1 generates due to the mass of box A m(A)
and Tension T2 arises due to mass of the system m(A) + m(B)
Thus tension T1 will be less than tension T2
T1 < T2
learn more about Tension force here:
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I assume you meant to say
Given that <em>x</em> = √3 and <em>x</em> = -√3 are roots of <em>f(x)</em>, this means that both <em>x</em> - √3 and <em>x</em> + √3, and hence their product <em>x</em> ² - 3, divides <em>f(x)</em> exactly and leaves no remainder.
Carry out the division:
To compute the quotient:
* 2<em>x</em> ⁴ = 2<em>x</em> ² • <em>x</em> ², and 2<em>x</em> ² (<em>x</em> ² - 3) = 2<em>x</em> ⁴ - 6<em>x</em> ²
Subtract this from the numerator to get a first remainder of
(2<em>x</em> ⁴ + 3<em>x</em> ³ - 5<em>x</em> ² - 9<em>x</em> - 3) - (2<em>x</em> ⁴ - 6<em>x</em> ²) = 3<em>x</em> ³ + <em>x</em> ² - 9<em>x</em> - 3
* 3<em>x</em> ³ = 3<em>x</em> • <em>x</em> ², and 3<em>x</em> (<em>x</em> ² - 3) = 3<em>x</em> ³ - 9<em>x</em>
Subtract this from the remainder to get a new remainder of
(3<em>x</em> ³ + <em>x</em> ² - 9<em>x</em> - 3) - (3<em>x</em> ³ - 9<em>x</em>) = <em>x</em> ² - 3
This last remainder is exactly divisible by <em>x</em> ² - 3, so we're left with 1. Putting everything together gives us the quotient,
2<em>x </em>² + 3<em>x</em> + 1
Factoring this result is easy:
2<em>x</em> ² + 3<em>x</em> + 1 = (2<em>x</em> + 1) (<em>x</em> + 1)
which has roots at <em>x</em> = -1/2 and <em>x</em> = -1, and these re the remaining zeroes of <em>f(x)</em>.