At Westonci.ca, we provide clear, reliable answers to all your questions. Join our vibrant community and get the solutions you need. Our platform provides a seamless experience for finding reliable answers from a knowledgeable network of professionals. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
Sagot :
To determine the length of a pendulum required for it to have a period of 1 second per cycle on the moon, where the acceleration due to gravity is \( g_{\text{moon}} = 1.6 \, \text{N/kg} \), we will use the formula that relates the period of a pendulum to its length and the acceleration due to gravity.
The formula for the period \( T \) of a simple pendulum is given by:
[tex]\[ T = 2\pi \sqrt{\frac{L}{g}} \][/tex]
Where:
- \( T \) is the period of the pendulum (in seconds).
- \( L \) is the length of the pendulum (in meters).
- \( g \) is the acceleration due to gravity (in N/kg).
We are given:
- The desired period \( T = 1 \) second.
- The acceleration due to gravity on the moon \( g = 1.6 \, \text{N/kg} \).
We need to calculate the length \( L \) for the pendulum to have this period. First, we solve for \( L \):
[tex]\[ T = 2\pi \sqrt{\frac{L}{g}} \][/tex]
We rearrange the formula to solve for \( L \):
[tex]\[ \sqrt{\frac{L}{g}} = \frac{T}{2\pi} \][/tex]
Square both sides to eliminate the square root:
[tex]\[ \frac{L}{g} = \left(\frac{T}{2\pi}\right)^2 \][/tex]
Multiply both sides by \( g \):
[tex]\[ L = g \left(\frac{T}{2\pi}\right)^2 \][/tex]
Substitute the given values \( T = 1 \, \text{s} \) and \( g = 1.6 \, \text{N/kg} \):
[tex]\[ L = 1.6 \left(\frac{1}{2\pi}\right)^2 \][/tex]
This simplifies to:
[tex]\[ L = 1.6 \left(\frac{1}{6.2832}\right)^2 \][/tex]
Calculating the fraction inside the parentheses:
[tex]\[ L = 1.6 \left(\frac{1}{6.2832}\right)^2 \approx 1.6 \times 0.02533 \approx 0.040528 \][/tex]
Thus, the length \( L \) of the pendulum must be approximately \( 0.040528 \) meters.
Comparing this result with the given choices:
- \( 3.2 \, \text{m} \)
- \( 7.9 \, \text{m} \)
- \( 0.25 \, \text{m} \)
- \( 0.041 \, \text{m} \)
The closest choice to our calculated length of \( 0.040528 \, \text{m} \) is:
[tex]\[ 0.041 \, \text{m} \][/tex]
Therefore, the correct length of the pendulum for it to have a period of 1 second per cycle on the moon is [tex]\( 0.041 \, \text{m} \)[/tex].
The formula for the period \( T \) of a simple pendulum is given by:
[tex]\[ T = 2\pi \sqrt{\frac{L}{g}} \][/tex]
Where:
- \( T \) is the period of the pendulum (in seconds).
- \( L \) is the length of the pendulum (in meters).
- \( g \) is the acceleration due to gravity (in N/kg).
We are given:
- The desired period \( T = 1 \) second.
- The acceleration due to gravity on the moon \( g = 1.6 \, \text{N/kg} \).
We need to calculate the length \( L \) for the pendulum to have this period. First, we solve for \( L \):
[tex]\[ T = 2\pi \sqrt{\frac{L}{g}} \][/tex]
We rearrange the formula to solve for \( L \):
[tex]\[ \sqrt{\frac{L}{g}} = \frac{T}{2\pi} \][/tex]
Square both sides to eliminate the square root:
[tex]\[ \frac{L}{g} = \left(\frac{T}{2\pi}\right)^2 \][/tex]
Multiply both sides by \( g \):
[tex]\[ L = g \left(\frac{T}{2\pi}\right)^2 \][/tex]
Substitute the given values \( T = 1 \, \text{s} \) and \( g = 1.6 \, \text{N/kg} \):
[tex]\[ L = 1.6 \left(\frac{1}{2\pi}\right)^2 \][/tex]
This simplifies to:
[tex]\[ L = 1.6 \left(\frac{1}{6.2832}\right)^2 \][/tex]
Calculating the fraction inside the parentheses:
[tex]\[ L = 1.6 \left(\frac{1}{6.2832}\right)^2 \approx 1.6 \times 0.02533 \approx 0.040528 \][/tex]
Thus, the length \( L \) of the pendulum must be approximately \( 0.040528 \) meters.
Comparing this result with the given choices:
- \( 3.2 \, \text{m} \)
- \( 7.9 \, \text{m} \)
- \( 0.25 \, \text{m} \)
- \( 0.041 \, \text{m} \)
The closest choice to our calculated length of \( 0.040528 \, \text{m} \) is:
[tex]\[ 0.041 \, \text{m} \][/tex]
Therefore, the correct length of the pendulum for it to have a period of 1 second per cycle on the moon is [tex]\( 0.041 \, \text{m} \)[/tex].
Thank you for choosing our service. We're dedicated to providing the best answers for all your questions. Visit us again. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.
