Answer:
786 Hz
Explanation:
Recall, the speed of sound is
v = 332 + 0.6t
Where t = 23°
v = 332 + 0.6(23)
v = 332 + 13.8
v = 345.8 m
Also, it is known that distance between two consecutive resonance length is half of the wavelength.
L2 - L1 = λ/2
34 - 12 = λ/2
λ/2 = 22
λ = 44 cm
Finally, remember that also
Frequency = speed/ wavelength
Frequency = 345.8/0.4
Frequency = 786 Hz
Therefore, the frequency of the tuning fork is 786 Hz
Wavelength = 20 m = 0.20 m
frequency = 3*10^8 / 0.20 = 1.5 * 10^9 Hertz
A) R = 1 / ( (1 / R1)+(1 / R2)+(1 / R3) )
b) I = U / R
c) I1 = U / R1 ... etc.
The coefficient of friction between the sled and the snow is 0.119.
To find the answer, we need to know about the friction.
<h3>How to find the coefficient of friction between the sled and the snow?</h3>
- Whenever a body moves over the surface of another body, a force come into play, which acts parallel to the surface of contact and oppose the relative motion. This opposing force is called friction.
- To solve the problem, we have to draw the free body diagram of the given system.
- We have given with the following values,
Here, acceleration will be equal to zero, because the velocity is given as constant.
- Thus, from the diagram, we can write the balancing equations as follows,
- Substituting N in f and f in the equation of ma, then we get,
- Substituting values, we get the coefficient of friction as,
Thus, we can conclude that, the coefficient of friction between the sled and the snow is 0.119.
Learn more about the friction here:
brainly.com/question/28107059
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A vaccum unlike sound,light can travel through any matter including a great vacuum of nothing (space)