Answer:
False the answer needs to be 20 characters :) good luck
Answer:
f = 409 Hz
Explanation:
We have,
Length of the open organ pipe, l = 0.29 m
Frequency of vibration of second overtone,
It is required to find the fundamental frequency of the pipe. For the open organ pipe, the frequency of second overtone is given by :
v is speed of sound
Let f is the fundamental frequency. It is given by :
The relation between f and f₂ can be written as :
So, the fundamental frequency of the pipe is 409 Hz.
Answer:
-0.01 mm
Explanation:
We are given that
The value of one division of vernier scale =0.5 mm
The value of one main scale division=0.49 mm
We have to find the value of least count of the instrument in mm.
We know that
Leas count of vernier caliper=1 main scale division-1 vernier scale division
Least count of vernier caliper=0.49-0.50=-0.01 mm
Hence, the least count of the instrument=-0.01 mm
Answer: -0.01 mm
The terminology is somewhat arbitrary, however, In this case, you intentionally alter the angle, which then determines the resulting distance. The presumption of the experiment is that distance is DEPENDENT on the angle. The dependent variable will be the how far the rock went. The dependent variable is the variable that you are testing. The first sentence of description indicates that the experiment is to determine launch Θ vs range.The largest range value is to be determined, so the range is plotted on the Y-axis. Angle Θ is plotted on the X-axis. The distance depends on the launch of the angle so the answer will be (D) how far the rock went. It is the dependent variable.