Answer:
5.571 sec
Explanation:
angular frequency = √ (k/m) = √ (49.3 / 5) = 3.14 rad/s
Period To = 2π / angular frequency
Period To = 2π/3.14 = 2 × 3.14 / 3.142 = 2.00 sec which you got
T measured by the observer = To / (√ (1 - (v²/c²))) = 2 / √( 1 - 0.871111) = 2 / 0.35901 = 5.571 sec
t=2.00/(1-√((2.80*10^8)^2/(3.00*10^8)^2))= should have been ( To / (√ (1 - (v²/c²))). where To = 2.00 sec
Because upward buoyant force is slightly higher than gravitation force for this particular object
The answer to the problem b.
Answer:
A)
B)
C)
Explanation:
Given that a pendulum is suspended by a shaft with a very light thin rod.
Followed by the given information: m = 100 g, I = 0.5 m, g = 9.8 m / s²
We can determine the answer to these questions using angular kinematics.
Angular kinematics is just derived from linear kinematics but in different symbols, and expressions.
Here are the formulas for angular kinematics:
- θ = ωt
- ∆w =
- L [Angular momentum] = mvr [mass × velocity × radius]
A) What is the minimum speed required for the pendulum to traverse the complete circle?
We can use the formula v = √gL derived from
B) The same question if the pendulum is suspended with a wire?
C) What is the ratio of the two calculated speeds?
The formula for average speed is

So we can just substitute our data.

- its the result