Answer:
From the data we know that runner A and runner B are 11 km apart from the start because (6+5) km
So the runner from the east direction has distance as unknown km, rate= 9 k/h ; time= d/r=x/9 hr
So runner towards the west will be
distance = 11-x, rate= 8 k/h, time = d/r = (11-x)/8
So equating east and west time we have
x/9= (11-x)/8
8x=99-9x
17x=99
x=5.92 km
That is the distance covered by runner towards the east and he will meet the runner toward the west at
6-5.92=0.08 km west of the flagpole.
Answer: 1.044 E -17 A
Explanation:
L =1.50m
A = 0.380mm = 3.8E- 7
Resistivity p =1.70 E-8
R =pl/A
R = 1.70 E-8 × 1.5/ 3.8 E-7
R = 6.71 E-16 ohms
V = IR
0.700 = I × 6.71 E-16
I =1.044E-17 A
<span>a.
the force in the direction of movement
Fx = F x cosA*
Fx = 12 x cos25*
Fx = 10.88
m x a = 10.88
a = 10.88/5
a= 2.18 m/s^2
b.
when the block will be lifted off the floor and when the vertical component
Fy = mg
Fy = F x sin25*
Fy= 5 x 9.8
Fy= 115.94 N
c.
if Fx = ma
a = 115.94/5
a= 23.19 m/s^2
hope it helps
</span>
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 !
