The length of a 2 sec pendulum is 1 m.
Given that, initial length of the simple pendulum L₁ = 1 m
Initial time period T₁ = 2 sec
We need to find the length of the pendulum whose time period is 2 sec
T₂ = 2 sec
L₂ = ?
We know that the time period of the simple pendulum is given by the formula,
T = 2π√(L/g)
From the above relation, we can write T ∝ √L
T₁ / T₂ = √(L₁/L₂)
Making L₂ from the above relation, we have,
L₂ = (T₂² * L₁)/ T₁² = 2² * 1/ 2² = 1 m
Thus, the length of a 2 sec pendulum is 1 m.
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Answer:
About 66 miles per hour
Explanation:
Based on the information given we can assume the car traveled the same number of miles every hour meaning all we need to do is divide.
400/6 ≈ 66 miles per hour
Answer: 250n
Explanation:
The formula for gravitational force is: F = (gMm)/r^2
There are two factors at play here:
1) The mass of the planet 'M'
2) The radius 'r'
We can ignore the small M and the g, they are constants that do not alter the outcome of this question.
You can see that both M and r are double that of earth. So lets say earth has M=1 and r=1. Then, new planet would have M=2 and r=2. Let's sub these two sets into the equation:
Earth. F = M/r^2 = 1/1
New planet. F = M/r^2 = 2/4 = 1/2
So you can see that the force on the new planet is half of that felt on Earth.
The question tells us that the force on earth is 500n for this person, so then on the new planet it would be half! So, 250n!
Quantitative because it tells you how many wheels