Ah hah ! That depends on where he weighs 1250N, because
weight depends not only on mass but also on the local gravity.
So the same mass can have different weights in different places,
and the same weight can indicate different mass in different places.
Weight = (mass) x (acceleration of gravity) .
Note: 1250N is almost exactly 281 pounds.
-- If he weighs 1250N on Mars, then his mass is 336.8 kg.
-- If he weighs 1250N on Earth, then his mass is 127.5 kg.
-- If he weighs 1250N on the Moon, then his mass is 770.2 kg.
This question is a big fat non sequitur !
The wavelength of radio waves traveling through vacuum only depends on the frequency that the radio station is licensed to broadcast on, (which had better be the frequency of the transmitter that they buy and use, or they're in big trouble).
The wavelength does NOT depend on the type of modulation that's used to put information onto the signal.
An amateur radio (ham) operator may very well start out using FM to talk over his radio to somebody else, and then for some reason they may decide to switch to AM. They can do that without ANY change in the wavelength of their transmissions.
Now, in the USA and many other countries, it so happens that all AM stations are licensed by their governments to transmit their programs on a channel somewhere between 500 KHz and 1.6 MHz, and all FM stations are licensed by their governments to transmit their programs on a channel somewhere between 88 MHz and 108 MHz. (And THAT's what the radio receivers in these countries are built to receive.)
Then we might say that all of the AM stations are grouped around 1 MHz, and all of the FM stations are grouped around 100 MHz. The FM frequencies are very roughly 100 times the AM frequencies, so the AM wavelengths are very roughly 100 times the FM wavelengths. That's <em>choice (3)</em> .
But please don't get the idea that it has anything to do with using AM or FM technology. It's just a matter of where in the spectrum the government decided to put the AM stations and where they put the FM stations.
For that matter . . . An analog TV station uses an AM signal for the picture and an FM signal for the sound, and it all goes in the same channel, with just about the same wavelengths !
Answer:
the shortest distance to the obstruction is 0.431 m
Explanation:
We can see this system as an air column, where the plumber is open and where the water is closed, in the case when he hears the sound there is a phenomenon of resonance and superposition of waves with constructive interference.
For the lowest resonance we must have a node where the water is and a maximum where the plumber is a quarter of the wavelength
λ = ¼ L
If we are in a major resonance specifically the following resonance. We have a full wavelength plus a quarter of the wavelength
λ = 4L / 3
The general formula is
λ = 4L / n n = 1, 3, 5, 7,…
In addition the wave speed is the product of the frequency by the wavelength
v = λ f
Let's replace
v = (4L / n) f
L = v n / (4 f)
Now we can calculate the depth or length of the air column
If we have the first standing wave n = 1
L = 340 1 / (4 197)
L = 0.431 m
If it is the second resonance n = 3
L = 340 3 / (4 197)
L = 1.29 m
We can see the shortest distance to the obstruction is 0.431 m
To solve this problem it is necessary to apply the second derivative of the function to find the maximum time reference and thus calculate the maximum voltage.
With the maximum voltage by Ohm's Law it is possible to find the maximum current.
Ohm's law defines that
E = I*R
Where,
I = Current
R= Resistance
On the other hand by faraday studies and the potential can be expressed at the rate of change of the electric flow, that is
Replacing with our values we have that
The second derivative is
When E' = 0 we have a Maximum, then
0 = -36t+36
t = 1
Therefore when the time is 1s E has a Maximum, replacing at the function
Then the maximum current will be given by
Therefore the maximum current induced in the ring is 6.42A
The answer C. 2p. I took the quiz and I got it correct.
