Again, looking this up:
resonant frequency[f] = 1 / (2 * pi * sqrt (LC))
original f = 1 / (2 * pi * sqrt (LC))
new f = 1 / (2 * pi * sqrt (2L * 2C))
= 1 / (2 * pi * 2 sqrt (LC))
= 1 / (4 * pi * sqrt (LC))
new f / original f = 1/4 / (1/2) = 1/2
new f = 1/2 * original f
So when you double the capacitance and double the inductance, the resonant frequency gets cut in half.
Originally posted by HolyTYes. What is your source, might I ask?>
Again, looking this up:
resonant frequency[f] = 1 / (2 * pi * sqrt (LC))
original f = 1 / (2 * pi * sqrt (LC))
new f = 1 / (2 * pi * sqrt (2L * 2C))
= 1 / (2 * pi * 2 sqrt (LC))
= 1 / (4 * pi * sqrt (LC))
new f / original f = 1/4 / (1/2) = 1/2
new f = 1/2 * original f
So when you double the capacitance and double the inductance, the resonant frequency gets cut in half.
In reflection, the light remains in the same medium, so its speed remains the same. In refraction, the speed changes. The change in speed is dependent on the wavelength of the light. The frequency remains the same. f[1] = f[2] = v[1]/wavelength[1] = v[2]/wavelength[2].
In its "desire" to find the quickest path from one point to another, the change in speed between the two mediums gives it a path that is not straight (it travels a greater distance in the faster medium and a shorter distance in the slower medium). The amount of change in the path (refraction) is dependent on the speed change which is dependent on the wavelength. In reflection, there's no change in speed, so the wavelength does not matter to how the light finds the quickest path.
Originally posted by HolyTYep, doing good.
In reflection, the light remains in the same medium, so its speed remains the same. In refraction, the speed changes. The change in speed is dependent on the wavelength of the light. The frequency remains the same. f[1] = f[2] = v[1]/wavelength[1] = v[2]/wavelength[2].
In its "desire" to find the quickest path from one point to another, the change in sp ...[text shortened]... change in speed, so the wavelength does not matter to how the light finds the quickest path.
Originally posted by RamnedNot quite sure what you mean by "the signs of the focal length and the object distance". But for the image distance:
Mirrors
[b]23. A concave mirror has a focal length of 15 cm. (A) If an object is placed 25 cm in front of the mirror, determine the signs of the focal length, the object distance, and the image distance. (B) Repeat part (A) if the object is placed 5 cm in front of the mirror.[/b]
The image distance (d_i), object distance (d_o) and focal length (f) are related by:
1/f = 1/d_i + 1/d_o
So (A) f = 15cm, d_o = 25cm => d_i = 37.5cm
(B) f = 15cm, d_o = 5cm => d_i = -7.5cm
(i.e. the image is a virtual image, appearing to be 7.5cm behind the mirror)
Originally posted by mtthwWhat I mean is - Positive or Negative for focal length and object distance...
Not quite sure what you mean by "the signs of the focal length and the object distance". But for the image distance:
The image distance (d_i), object distance (d_o) and focal length (f) are related by:
1/f = 1/d_i + 1/d_o
So (A) f = 15cm, d_o = 25cm => d_i = 37.5cm
(B) f = 15cm, ...[text shortened]... d_i[/i] = -7.5cm
(i.e. the image is a virtual image, appearing to be 7.5cm behind the mirror)
Originally posted by PBE6Then it'll land in the lake and hit the dock and the boat. The other side is a forest so it will just sit up against a tree.
Chop it down from the other side! 😵
I know it has something to do with measuring the angle from a set distance from the base of the tree but I forget whether it's SIN COS or TAN
Originally posted by uzlessMeasure the shadow.
Then it'll land in the lake and hit the dock and the boat. The other side is a forest so it will just sit up against a tree.
I know it has something to do with measuring the angle from a set distance from the base of the tree but I forget whether it's SIN COS or TAN
At the same time measure the shadow of a metre high stick.
Then just scale up.
Your answer in metres is tree height = tree shadow divided by stick shadow.
Originally posted by wolfgang59That's an excellent Idea except the sun is always shining into the forest instead of behind it.
Measure the shadow.
At the same time measure the shadow of a metre high stick.
Then just scale up.
Your answer in metres is tree height = tree shadow divided by stick shadow.
PB6 told me the Tan equation but I like your Ratio idea. So much simpler.
Thanks