Sure, let's break down both parts of the question step-by-step.
### Problem 10: Make [tex]\( w \)[/tex] the subject of the formula
We are given the equation:
[tex]\[ t = 2 - \frac{3w}{a} \][/tex]
Our goal is to solve this equation for [tex]\( w \)[/tex].
1. Isolate the term involving [tex]\( w \)[/tex]:
[tex]\[ t - 2 = -\frac{3w}{a} \][/tex]
2. Multiply both sides by [tex]\( -1 \)[/tex] to remove the negative sign:
[tex]\[ -(t - 2) = \frac{3w}{a} \][/tex]
Simplifying, we get:
[tex]\[ 2 - t = \frac{3w}{a} \][/tex]
3. Multiply both sides by [tex]\( a \)[/tex] to get rid of the fraction:
[tex]\[ a(2 - t) = 3w \][/tex]
4. Divide both sides by 3 to solve for [tex]\( w \)[/tex]:
[tex]\[ w = \frac{a(2 - t)}{3} \][/tex]
So, the solution is:
[tex]\[ w = \frac{a(2 - t)}{3} \][/tex]
### Problem 11: Periodic time [tex]\( T \)[/tex] of a pendulum
We know that the periodic time [tex]\( T \)[/tex] varies directly as the square root of its length [tex]\( l \)[/tex]. This can be expressed as:
[tex]\[ T \propto \sqrt{l} \][/tex]
[tex]\[ T = k \sqrt{l} \][/tex]
where [tex]\( k \)[/tex] is the proportionality constant.
We are given:
[tex]\[ T = 6 \text{ when } l = 9 \][/tex]
1. Find the proportionality constant [tex]\( k \)[/tex]:
[tex]\[ 6 = k \sqrt{9} \][/tex]
Simplifying, we get:
[tex]\[ 6 = k \times 3 \][/tex]
Therefore,
[tex]\[ k = \frac{6}{3} = 2 \][/tex]
2. Use [tex]\( k \)[/tex] to find the new [tex]\( T \)[/tex] when [tex]\( l = 25 \)[/tex]:
[tex]\[ T = k \sqrt{25} \][/tex]
Substituting [tex]\( k = 2 \)[/tex], we get:
[tex]\[ T = 2 \times \sqrt{25} \][/tex]
Simplifying, we have:
[tex]\[ T = 2 \times 5 = 10 \][/tex]
So, the new periodic time [tex]\( T \)[/tex] when [tex]\( l = 25 \)[/tex] is:
[tex]\[ T = 10 \][/tex]
To summarize:
- For Problem 10, [tex]\( w = \frac{a(2 - t)}{3} \)[/tex]
- For Problem 11, [tex]\( T = 10 \)[/tex]
I hope this helps! Let me know if you have any further questions.
