Answer:
Explanation:
In order to answer this problem you have to know the depth of the column, we say R, this information is important because allows you to compute some harmonic of the tube. With this information you can compute the depth of the colum of air, by taking tino account that the new depth is R-L.
To find the fundamental mode you use:
n: mode of the sound
vs: sound speed
L: length of the column of air in the tube.
A) The fundamental mode id obtained for n=1:
B) For the 3rd harmonic you have:
C) For the 2nd harmonic:
Power is the rate work done given by dividing work done by unit time. It is measured in watts equivalent to J/s.
In this case the force by the student is mg = 490 N (taking g as 9.8m/s²)
Work done is given by force × distance,
Therefore, Power =(force × distance)/ time, but velocity/speed =distance/time
Thus, Power = force × speed/velocity
= 490 N × 1.25
= 612.5 J/S (Watts)
Hence, power will be 612.5 Watts.
2.) <span>Heat is transferred from your body into an ice cube by "Conduction"
3.) </span><span>The loudness of a sound will be determined by the wave’s "Frequency"
4.) </span>The speed of a sound wave <span>"is slower than an electromagnetic wave"
Hope this helps!
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Answer:
e = 1.21 mV
Explanation:
given,
length of rod = 10 m
height of drop = 4.89 m
Earth’s magnetic field = 12.4 µT
acceleration of gravity = 9.8 m/s²
velocity of the beam
v = 9.79 m/s
emf of the beam
e = B l v
e = 12.4 x 10⁻⁶ x 9.79 x 10
e = 1.21 x 10⁻³ V
e = 1.21 mV
Answer:
cold cathode fluorescent lamp
Explanation:
A cold cathode fluorescent lamp (CCFL) is a lighting system that uses two phenomena: electron discharge and fluorescence.
CCFLs are mainly used as light sources for backlights, since they are compact and durable than ordinary fluorescent lamps. They exhibit a wide range of brightness and color (color temperature and chromaticity) that can be achieved by varying the pressure and type of the material injected into the glass tube. The thickness and type of the phosphor used to coat the inner wall of the tube also plays a role in altering the color and brightness.
