Given:
F = ax
where
x = distance by which the rubber band is stretched
a = constant
The work done in stretching the rubber band from x = 0 to x = L is
Answer:
Consider that the bar magnet has a magnetic field that is acting around it, which will imply that there is a change in the magnetic flux through the loop whenever it moves towards the conducting loop. This could be described as an induction of the electromotive Force in the circuit from Faraday's law.
In turn by Lenz's law, said electromotive force opposes the change in the magnetic flux of the circuit. Therefore, there is a force that opposes the movement of the bar magnet through the conductor loop. Therefore, the bar magnet does not suffer free fall motion.
The bar magnet does not move as a freely falling object.
Does this help?
When an object is
immersed in a fluid (in this case water, but may include both liquids and
gases) the fluid exerts an upward force on the object which is called buoyancy
force or <span>up-thrust. Archimedes’ Principle states that the buoyant
force (upward push or force) applied to an object is equal to the weight of the fluid that the object takes the space of by
that object. Thus when an object is
placed in water the rise in the water level is dictated by the mass of that
object.</span>
<span>
</span>
<span>So for example if you fill a bucket with water and you drop a stone in that bucket, if you measure the weight of the water that overflows from the bucket due to the stone being dropped into the bucket is equivalent to the pushing force that the water has on the stone (as the stone drops to the bottom of the bucket the water is pushing it to stay afloat but the rock is more dense than water and as such its downthrust exceeds water's upthrust).</span>
A camera flash or lighting bolt because stored separated positive and negative charges --> caused them to do work by briefly lighting a bulb as the separated charges moved back together
Answer:
fb = 240.35 Hz
Explanation:
In order to calculate the beat frequency generated by the first modes of each, organ and tube, you use the following formulas for the fundamental frequencies.
Open tube:
(1)
vs: speed of sound = 343m/s
L: length of the open tube = 0.47328m
You replace in the equation (1):
Closed tube:
L': length of the closed tube = 0.702821m
Next, you use the following formula for the beat frequency:
The beat frequency generated by the first overtone pf the closed pipe and the fundamental of the open pipe is 240.35Hz