Answer:
Option A is correct - While a guitar string is vibrating, you gently touch the midpoint of the string to ensure that the string does not vibrate at that point. The lowest-frequency standing wave that could be present on the string vibrates at twice the fundamental frequency.
Explanation:
Before touching the midpoint of the string, the string vibrates with one loop.
Fundamental frequency, f1 = v/(2*L)
Now, when the midpoint of the guitar string was touched, the string vibrates with two loops.
Hence, f2 = 2*v/(2*L)
f2 = 2*f1
Therefore, compared to the fundamental frequency the frequency would be double.
Option A is correct - While a guitar string is vibrating, you gently touch the midpoint of the string to ensure that the string does not vibrate at that point. The lowest-frequency standing wave that could be present on the string vibrates at twice the fundamental frequency.
Answer: be found in group 17 and be highly reactive
Explanation:
Elements are distributed in groups and periods in a periodic table.
Elements that belong to same groups will show similar chemical properties because they have same number of valence electrons.
Flourine, chlorine, bromine and iodine are elements which belong to Group 17. All of them contain 7 valence electrons each and need one electron to complete their octet.
The chemical reactivity of elements is governed by the valence electrons present in the element and thus all of them are highly reactive.
Answer: 1500metres
Explanation: the formula for distance is speed × time which equals 150m/min × 10 min = 1500m
This formula can be rearranged...
Speed = Distance ÷ time.
Time= Distance ÷ speed.
Distance= Time × speed.
Answer:
The magnetic field exerts net force and net torque on the loop.
