Answer :
Sure, let's solve for [tex]\( g \)[/tex] in the equation [tex]\( T = 2\pi \sqrt{\frac{L}{g}} \)[/tex] given that [tex]\( T = 1.57 \)[/tex] and [tex]\( L = 2 \)[/tex].
1. Rewrite the formula:
[tex]\[
T = 2\pi \sqrt{\frac{L}{g}}
\][/tex]
2. Isolate [tex]\( g \)[/tex]:
First, square both sides of the equation to get rid of the square root:
[tex]\[
T^2 = (2\pi)^2 \left(\frac{L}{g}\right)
\][/tex]
3. Simplify the expression:
[tex]\[
T^2 = 4\pi^2 \left(\frac{L}{g}\right)
\][/tex]
4. Rearrange to solve for [tex]\( g \)[/tex]:
[tex]\[
g = \frac{4\pi^2 L}{T^2}
\][/tex]
5. Plug in the given values:
[tex]\[
g = \frac{4\pi^2 \cdot 2}{1.57^2}
\][/tex]
6. Calculate intermediate values:
- Calculate [tex]\( (2 \pi)^2 \)[/tex]:
[tex]\[
(2\pi)^2 = 4\pi^2 \approx 39.4784
\][/tex]
- Calculate [tex]\( 1.57^2 \)[/tex]:
[tex]\[
1.57^2 \approx 2.4649
\][/tex]
7. Finally, calculate [tex]\( g \)[/tex]:
[tex]\[
g = \frac{39.4784 \cdot 2}{2.4649} \approx 32.0325
\][/tex]
So, the value of [tex]\( g \)[/tex] is approximately [tex]\( 32.0325 \)[/tex].
1. Rewrite the formula:
[tex]\[
T = 2\pi \sqrt{\frac{L}{g}}
\][/tex]
2. Isolate [tex]\( g \)[/tex]:
First, square both sides of the equation to get rid of the square root:
[tex]\[
T^2 = (2\pi)^2 \left(\frac{L}{g}\right)
\][/tex]
3. Simplify the expression:
[tex]\[
T^2 = 4\pi^2 \left(\frac{L}{g}\right)
\][/tex]
4. Rearrange to solve for [tex]\( g \)[/tex]:
[tex]\[
g = \frac{4\pi^2 L}{T^2}
\][/tex]
5. Plug in the given values:
[tex]\[
g = \frac{4\pi^2 \cdot 2}{1.57^2}
\][/tex]
6. Calculate intermediate values:
- Calculate [tex]\( (2 \pi)^2 \)[/tex]:
[tex]\[
(2\pi)^2 = 4\pi^2 \approx 39.4784
\][/tex]
- Calculate [tex]\( 1.57^2 \)[/tex]:
[tex]\[
1.57^2 \approx 2.4649
\][/tex]
7. Finally, calculate [tex]\( g \)[/tex]:
[tex]\[
g = \frac{39.4784 \cdot 2}{2.4649} \approx 32.0325
\][/tex]
So, the value of [tex]\( g \)[/tex] is approximately [tex]\( 32.0325 \)[/tex].
