Hi there! Lets see!
- m is mass, and its units are kg
- k is the elastic constant measured in newtons per meter (N/m), or kilograms per second squared kg/s²
Therefore:
![\sqrt{\dfrac{m}{k}} =\sqrt{\dfrac{[kg]}{[\dfrac{kg}{s^2}]}} =\sqrt{\dfrac{[kg]}{[kg]}\cdot s^2} = \sqrt{[s]^2} = s](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cdfrac%7Bm%7D%7Bk%7D%7D%20%3D%5Csqrt%7B%5Cdfrac%7B%5Bkg%5D%7D%7B%5B%5Cdfrac%7Bkg%7D%7Bs%5E2%7D%5D%7D%7D%20%20%3D%5Csqrt%7B%5Cdfrac%7B%5Bkg%5D%7D%7B%5Bkg%5D%7D%5Ccdot%20s%5E2%7D%20%3D%20%5Csqrt%7B%5Bs%5D%5E2%7D%20%3D%20s)
The period is given in seconds so the formula is dimensionally correct.
I think it's 39.53
(please do calculate it. I am not completely sure)
Answer:
1. 77.31 N/m
2. 26.2 m/s
3. increase
Explanation:
1. According to the law of energy conservation, when she jumps from the bridge to the point of maximum stretch, her potential energy would be converted to elastics energy. Her kinetic energy at both of those points are 0 as speed at those points are 0.
Let g = 9.8 m/s2. And the point where the bungee ropes are stretched to maximum be ground 0 for potential energy. We have the following energy conservation equation


where m = 75 kg is the mass of the jumper, h = 72 m is the vertical height from the jumping point to the lowest point, k (N/m) is the spring constant and x = 72 - 35 = 37 m is the length that the cord is stretched


2. At 35 m below the platform, the cord isn't stretched, so there isn't any elastics energy, only potential energy converted to kinetics energy. This time let's use the 35m point as ground 0 for potential energy

where H = 35m this time due to the height difference between the jumping point and the point 35m below the platform


3. If she jumps from her platform with a velocity, then her starting kinetic energy is no longer 0. The energy conservation equation would then be

So the elastics energy would increase, which would lengthen the maximum displacement of the cord
Answer:
B: It is moving north.
Explanation:
We are told that a train is moving with a velocity of 70 km/h.
Now, we are also told that the positive direction is north.
This means that the train is moving in the northern direction.
We can't say if it is speeding up or not because we are not given the acceleration.
Thus, option B is correct