The current is defined as the amount of charge transferred through a certain point in a certain time interval:
where
I is the current
Q is the charge
is the time interval
For the lightning bolt in our problem, Q=6.0 C and
, so the average current during the event is
My response to question (a) and (b) is that all of the element of the rope need to aid or support the weight of the rope and as such, the tension will tend to increase along with height.
Note that It increases linearly, if the rope is one that do not stretch. So, the wave speed v= √ T/μ increases with height.
<h3>How does tension affect the speed of a wave in a rope?</h3>
The Increase of the tension placed on a string is one that tends to increases the speed of a wave, which in turn also increases the frequency of any given length.
Therefore, My response to question (a) and (b) is that all of the element of the rope need to aid or support the weight of the rope and as such, the tension will tend to increase along with height. Note that It increases linearly, if the rope is one that do not stretch. So, the wave speed v= √ T/μ increases with height.
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See full question below
(a) If a long rope is hung from a ceiling and waves are sent up the rope from its lower end, why does the speed of the waves change as they ascend? (b) Does the speed of the ascending waves increase or decrease? Explain.
I am using the equation F=ma (force equals mass times acceleration) to solve these problems.
1. You are looking for force, and have mass and acceleration. You just plug in the values for mass and acceleration to get the force needed.
F=(15kg)(5m/s^2)
F=75N
2. Again, you are looking for force, and just need to plug in the values for mass and acceleration
F=(3kg)(2.4m/s^2)
F=7.2N
3. In this problem, you have force and mass, but need to find acceleration. To do this, you need to get acceleration alone on one side of the equation - divide each side by m. Your equation will now be F/m=a
a=(5N)/(3.7kg)
a=18.5m/s^2
I did not use significant figures. Let me know if you need to do that and need any help on that. Hope this helps!
